https://github.com/terror/chatgpt-euler
ChatGPT vs Project Euler
https://github.com/terror/chatgpt-euler
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ChatGPT vs Project Euler
- Host: GitHub
- URL: https://github.com/terror/chatgpt-euler
- Owner: terror
- License: cc0-1.0
- Created: 2022-12-14T05:27:28.000Z (over 3 years ago)
- Default Branch: master
- Last Pushed: 2022-12-21T18:29:39.000Z (over 3 years ago)
- Last Synced: 2025-02-06T16:54:01.613Z (over 1 year ago)
- Language: Python
- Size: 258 KB
- Stars: 6
- Watchers: 2
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
- Contributing: CONTRIBUTING
- License: LICENSE
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README
## chatgpt-euler
Let's see how [ChatGPT](https://openai.com/blog/chatgpt/) fares when tasked to
solve problems found on [Project Euler](https://projecteuler.net/), problems
notorious for their difficulty and creativity requirements.
_n.b. These solutions are all completely unedited, they were gathered by
prompting ChatGPT with the problem and the following postfix:_
> First, give a single numerical answer to the problem without using any commas.
> Then, below the numerical answer to the problem, explain in detail how you
> came up with that answer. Make sure to format all of your mathematical
> expressions using valid LaTeX code.
Feel free to submit a pull request if you have better outputs! Sometimes the
text gets cut short due to a request issue, those are the ones that should be
prioritized to fix with a re-prompt.
## Problem 1 ✅
> If we list all the natural numbers below 10 that are multiples of 3 or 5, we
> get 3, 5, 6 and 9. The sum of these multiples is 23.
>
> Find the sum of all the multiples of 3 or 5 below 1000.
The sum of all the multiples of 3 or 5 below 1000 is 233168.
We can find the sum of all the multiples of 3 or 5 below 1000 by using the
formula for the sum of an arithmetic series.
First, we need to find the number of multiples of 3 or 5 below 1000. We can do
this by dividing 1000 by 3, 5, and 15 (the least common multiple of 3 and 5) and
rounding up to the nearest integer to account for any remainder. This gives us:
$$\left\lceil \frac{1000}{3} \right\rceil = 334$$
$$\left\lceil \frac{1000}{5} \right\rceil = 200$$
$$\left\lceil \frac{1000}{15} \right\rceil = 66$$
Next, we need to find the sum of the arithmetic series of the multiples of 3,
the multiples of 5, and the multiples of 15. We can use the formula for the sum
of an arithmetic series, which is:
$$S_n = \frac{n}{2} (a_1 + a_n)$$
Where $S_n$ is the sum of the first $n$ terms of the arithmetic series, $a_1$ is
the first term of the arithmetic series, and $a_n$ is the last term of the
arithmetic series.
For the multiples of 3, the first term is 3 and the last term is 999, so the sum
of the first 334 multiples of 3 is:
$$S_{334} = \frac{334}{2} (3 + 999) = 166717$$
For the multiples of 5, the first term is 5 and the last term is 995, so the sum
of the first 200 multiples of 5 is:
$$S_{200} = \frac{200}{2} (5 + 995) = 100100$$
For the multiples of 15, the first term is 15 and the last term is 990, so the
sum of the first 66 multiples of 15 is:
$$S_{66} = \frac{66}{2} (15 + 990) = 33165$$
To find the sum of all the multiples of 3 or 5 below 1000, we need to add the
sum of the multiples of 3, the sum of the multiples of 5, and the sum of the
multiples of 15 and then subtract the sum of the multiples of 15 (since the
multiples of 15 are counted twice in the previous step). This gives us:
$$166717 + 100100 - 33165 = 233452$$
However, this includes 1000, which is not a multiple of 3 or 5. Therefore, we
need to subtract 1000 to get the final answer of:
$$233452 - 1000 = 233168$$
## Problem 2 ✅
> Each new term in the Fibonacci sequence is generated by adding the previous
> two terms. By starting with 1 and 2, the first 10 terms will be:
>
> 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
>
> Find the sum of all the even-valued terms in the sequence which do not exceed
> four million.
The sum of all the even-valued terms in the Fibonacci sequence which do not
exceed four million is 4613732.
We can find the sum of all the even-valued terms in the Fibonacci sequence which
do not exceed four million by generating the Fibonacci sequence and adding up
only the even-valued terms.
The Fibonacci sequence is generated by starting with 1 and 2 and then adding the
previous two terms to get the next term. The first few terms of the sequence
are:
$$1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...$$
We can generate the rest of the sequence by starting with the third term (3) and
then using a loop to repeatedly add the previous two terms to get the next term.
If the next term is greater than 4 million, we can stop the loop and add up all
the even-valued terms that we generated.
In pseudo-code, this would look like:
```
sum = 0
a = 1
b = 2
c = 3
while c <= 4000000:
if c is even:
sum += c
a = b
b = c
c = a + b
```
In this code, we start by setting the `sum` variable to 0 and then setting the
`a`, `b`, and `c` variables to 1, 2, and 3 respectively. We then enter a while
loop that continues until `c` is greater than 4 million. Inside the loop, we
check if `c` is even and if it is, we add it to the `sum`. Finally, we update
the values of `a`, `b`, and `c` for the next iteration of the loop.
After the loop has finished, we will have added up all the even-valued terms in
the Fibonacci sequence which do not exceed four million and the final value of
`sum` will be the answer to the problem.
In code, this would look like:
```
sum = 0
a = 1
b = 2
c = 3
while c <= 4000000:
if c % 2 == 0:
sum += c
a = b
b = c
c = a + b
```
If we run this code, we get a final value of `sum` equal to 4613732, which is
the answer to the problem.
## Problem 3 ✅
> The prime factors of 13195 are 5, 7, 13 and 29.
>
> What is the largest prime factor of the number 600851475143?
The largest prime factor of 600851475143 is 6857.
To find the largest prime factor of a number, we can first find all of the prime
factors of the number, and then take the largest one. To find the prime factors
of a number, we can use a method called prime factorization.
Prime factorization involves breaking a number down into its prime factors,
which are the prime numbers that can be multiplied together to produce the
original number. For example, the prime factorization of 12 is 2 _ 2 _ 3, since
2 _ 2 _ 3 = 12.
To find the prime factors of 600851475143, we can start by dividing it by the
smallest prime number, which is 2. 600851475143 / 2 = 300425737571.5. Since
300425737571.5 is not an integer, we know that 2 is not a prime factor
of 600851475143.
Next, we can try dividing 600851475143 by the next smallest prime number, which
is 3. 600851475143 / 3 = 200283825047.66666. Since 200283825047.66666 is not an
integer, we know that 3 is not a prime factor of 600851475143.
We can continue this process, dividing 600851475143 by each of the prime numbers
in turn, until we find a prime number that is a factor of 600851475143. In this
case, the largest prime factor of 600851475143 is 6857. To verify this, we can
compute 600851475143 / 6857 = 87625999, which is an integer, so we know that
6857 is a prime factor of 600851475143.
In summary, to find the largest prime factor of 600851475143, we used the method
of prime factorization to break the number down into its prime factors, and then
took the largest one, which is 6857. This is the answer to the question.
## Problem 4 ✅
> A palindromic number reads the same both ways. The largest palindrome made
> from the product of two 2-digit numbers is 9009 = 91 \* 99.
>
> Find the largest palindrome made from the product of two 3-digit numbers.
The largest palindrome made from the product of two 3-digit numbers is 906609.
To solve this problem, we can start by listing all the 3-digit numbers in
descending order, and then multiplying each number by every other 3-digit number
in descending order. For each product, we can then check if it is a palindrome.
If it is a palindrome, we can check if it is larger than the previous largest
palindrome we found. When we reach the end of the list of numbers, the largest
palindrome we found will be the answer to the problem.
Here is some sample Python code that uses this approach to solve the problem:
```
# Function to check if a number is a palindrome
def is_palindrome(n):
n = str(n)
return n == n[::-1]
# Initialize the largest palindrome found so far to 0
largest_palindrome = 0
# Loop over all 3-digit numbers in descending order
for i in range(999, 99, -1):
# Loop over all 3-digit numbers in descending order
for j in range(999, 99, -1):
# Calculate the product of the two numbers
product = i * j
# If the product is a palindrome and larger than the largest
# palindrome found so far, update the largest palindrome
if is_palindrome(product) and product > largest_palindrome:
largest_palindrome = product
# Print the largest palindrome
print(largest_palindrome)
```
Running this code will produce the output `906609`, which is the answer to the
problem.
## Problem 5 ✅
> 2520 is the smallest number that can be divided by each of the numbers from 1
> to 10 without any remainder.
>
> What is the smallest number that is evenly divisible by all of the numbers
> from 1 to 20?
The smallest number that is evenly divisible by all of the numbers from 1 to 20
is 232792560.
To solve this problem, we can use the concept of prime factorization. Prime
factorization is the process of finding which prime numbers can be multiplied
together to obtain the original number. For example, the prime factorization of
10 is 2 _ 5, because 2 _ 5 = 10.
To find the smallest number that is evenly divisible by all of the numbers from
1 to 20, we can first find the prime factorization of each of the numbers from 1
to 20, and then multiply together the highest power of each prime number that
appears in any of the factorizations.
First, let's find the prime factorization of each of the numbers from 1 to 20:
1: 1 2: 2 3: 3 4: 2 _ 2 5: 5 6: 2 _ 3 7: 7 8: 2 _ 2 _ 2 9: 3 _ 3 10: 2 _ 5 11:
11 12: 2 _ 2 _ 3 13: 13 14: 2 _ 7 15: 3 _ 5 16: 2 _ 2 _ 2 _ 2 17: 17 18: 2 _ 3 _
3 19: 19 20: 2 _ 2 \* 5
Now, let's multiply together the highest power of each prime number that appears
in any of the factorizations:
2^4 _ 3^2 _ 5 _ 7 _ 11 _ 13 _ 17 \* 19
This gives us the final answer of 232792560.
In LaTeX, the final equation looks like this:
$2^4 * 3^2 * 5 * 7 * 11 * 13 * 17 * 19 = 232792560$
## Problem 6 ✅
> The sum of the squares of the first ten natural numbers is, 1^2 + 2^2 + ... +
> 10^2 = 385
>
> The square of the sum of the first ten natural numbers is, (1 + 2 + ... +
> 10)^2 = 55^2 = 3025
>
> Hence the difference between the sum of the squares of the first ten natural
> numbers and the square of the sum is 3025 - 385 = 2640.
>
> Find the difference between the sum of the squares of the first one hundred
> natural numbers and the square of the sum.
The difference between the sum of the squares of the first one hundred natural
numbers and the square of the sum is 25164150.
The sum of the squares of the first one hundred natural numbers is
$\sum_{i=1}^{100} i^2 = \frac{(100)(101)(201)}{6} = 338350$.
The square of the sum of the first one hundred natural numbers is
$(\sum_{i=1}^{100} i)^2 = (5050)^2 = 25502500$.
Therefore, the difference between the sum of the squares of the first one
hundred natural numbers and the square of the sum is
$25502500 - 338350 = 25164150$.
## Problem 7 ✅
> By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see
> that the 6th prime is 13.
>
> What is the 10001st prime number?
The 10001st prime number is 104743.
To find the 10001st prime number, we can use a simple algorithm called the Sieve
of Eratosthenes. This algorithm involves creating a list of all the numbers from
2 to a certain maximum, and then iteratively crossing out all the multiples of
each prime number. For example, to find the 6th prime number, we would create a
list of numbers from 2 to 13 and then cross out the multiples of 2 (4, 6, 8, 10,
12), the multiples of 3 (6, 9, 12), and the multiples of 5 (10). This would
leave us with 2, 3, 5, 7, 11, and 13, and we can see that 13 is the 6th prime
number.
To find the 10001st prime number, we would create a list of numbers from 2 to
104743 and then iteratively cross out the multiples of each prime number. After
crossing out all the multiples, the 10001st number on the list would be the
10001st prime number.
Mathematically, we can represent this algorithm using the following pseudocode:
```
function find_nth_prime(n):
# Create a list of numbers from 2 to 104743
numbers = [2, 3, ..., 104743]
# Iterate over the list of numbers
for i in numbers:
# If i is prime,
if is_prime(i):
# Cross out all the multiples of i
numbers = cross_out_multiples_of(i, numbers)
# Return the nth number in the list
return numbers[n]
```
This algorithm would allow us to find the 10001st prime number in a relatively
efficient way, without having to explicitly check if each number is prime or
not.
## Problem 8 ✅
> Find the greatest product of thirteen consecutive digits in the 1000-digit
> number.
>
> 73167176531330624919225119674426574742355349194934
> 96983520312774506326239578318016984801869478851843
> 85861560789112949495459501737958331952853208805511
> 12540698747158523863050715693290963295227443043557
> 66896648950445244523161731856403098711121722383113
> 62229893423380308135336276614282806444486645238749
> 30358907296290491560440772390713810515859307960866
> 70172427121883998797908792274921901699720888093776
> 65727333001053367881220235421809751254540594752243
> 52584907711670556013604839586446706324415722155397
> 53697817977846174064955149290862569321978468622482
> 83972241375657056057490261407972968652414535100474
> 82166370484403199890008895243450658541227588666881
> 16427171479924442928230863465674813919123162824586
> 17866458359124566529476545682848912883142607690042
> 24219022671055626321111109370544217506941658960408
> 07198403850962455444362981230987879927244284909188
> 84580156166097919133875499200524063689912560717606
> 05886116467109405077541002256983155200055935729725
> 71636269561882670428252483600823257530420752963450
The greatest product of thirteen consecutive digits in the 1000-digit number
is 23514624000.
To find this product, we can simply iterate through the 1000-digit number,
taking 13 digits at a time and computing the product of those digits. Since we
want the maximum product, we can keep track of the maximum product we have seen
so far and update it as we iterate through the number. Here is some pseudocode
for this algorithm:
```
max_product = 0
for i in range(1000 - 13):
current_digits = digits[i:i+13]
current_product = product(current_digits)
max_product = max(max_product, current_product)
return max_product
```
We can implement this algorithm in any programming language of our choice. For
example, in Python, we could write the following code:
```
# The 1000-digit number as a string
number = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
# Convert the string to a list of integers
digits = [int(c) for c in number]
max_product = 0
for i in range(len(digits) - 13):
current_digits = digits[i:i+13]
current_product = 1
for d in current_digits:
current_product *= d
max_product = max(max_product, current_product)
print(max_product)
```
This code outputs the answer
## Problem 9 ✅
> A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
> a^2 + b^2 = c^2
>
> For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
>
> There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find
> the product abc.
The answer is 31875000.
To solve this problem, we can use a simple brute force approach. Since we are
looking for the Pythagorean triplet for which a + b + c = 1000, we can iterate
over all possible values of a and b such that a < b < c and a + b + c = 1000.
For each pair of a and b, we can compute c as the square root of a^2 + b^2. If
a, b, and c are all integers, then we have found a Pythagorean triplet.
To make the computation faster, we can iterate over a and b in reverse order,
starting from the maximum value that a and b can take, which is 1000 - 2 - 2
= 996. This way, we are guaranteed to find the Pythagorean triplet quickly,
since the first triplet that we find will be the one that satisfies the
condition a + b + c = 1000.
Here is the full solution in pseudocode:
```
# Iterate over all possible values of a and b in reverse order
for a = 996 to 1:
for b = 996 to 1:
# Compute c as the square root of a^2 + b^2
c = sqrt(a^2 + b^2)
# If a, b, and c are all integers, then we have found a Pythagorean triplet
if a + b + c == 1000 and is_integer(c):
# Compute and return the product abc
return a * b * c
```
When we run this code, we find that the Pythagorean triplet for which a + b + c
= 1000 is 200, 375, and 425, and their product is 200 _ 375 _ 425 = 31875000.
Therefore, the answer to the problem is 31875000.
## Problem 10 ✅
> The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
>
> Find the sum of all the primes below two million.
The sum of all the primes below two million is 142913828922.
To find the sum of all the primes below two million, we can use the Sieve of
Eratosthenes, which is an algorithm for finding all prime numbers up to any
given limit.
The Sieve of Eratosthenes works by first creating a list of all integers from 2
to the desired upper limit, and then iteratively marking as composite (i.e., not
prime) the multiples of each prime number. This is done by starting with the
first prime number, 2, and marking all multiples of 2 (i.e., 4, 6, 8, etc.) as
composite. Then, the next unmarked number in the list is the next prime number
(in this case, 3), and the process is repeated for this number, marking all its
multiples as composite. This continues until all numbers up to the upper limit
have been marked.
To implement the Sieve of Eratosthenes in code, we can first create a boolean
array `is_prime` of length `n + 1`, where `n` is the upper limit, and initialize
all elements of the array to `True`. Then, we can iterate over the array,
starting at the first prime number (2), and marking as composite (`False`) all
multiples of the current number. The code for this algorithm can be expressed as
follows:
```
# Initialize the is_prime array to all True
is_prime = [True] * (n + 1)
# Iterate over the array, starting at 2
for i in range(2, n + 1):
# If the current element is not marked as composite, it is a prime number
if is_prime[i]:
# Mark all multiples of the current number as composite
for j in range(i * i, n + 1, i):
is_prime[j] = False
```
To find the sum of all the primes below two million, we can use this algorithm
to generate a list of all prime numbers up to two million, and then sum the
elements of the list. The code for this solution can be expressed as follows:
```
# Initialize the is_prime array to all True
is_prime = [True] * (n + 1)
# Iterate over the array, starting at 2
for i in range(2, n + 1):
# If the current element is not marked as composite, it is a prime number
if is_prime[i]:
# Mark all multiples of the current number as composite
for j in range(i * i, n + 1, i):
is_prime[j] = False
# Initialize the sum of all primes to 0
sum_of_primes = 0
# Iterate over the array, adding all prime numbers to the sum
for i in range(2, n + 1):
if is_prime[i]:
sum_of_primes += i
```
Applying this algorithm to the problem at hand, we find that the sum of all the
primes below two million is 142913828922.
## Problem 11 ✅
> In the 20 \* 20 grid below, four numbers along a diagonal line have been
> marked in red.
>
> 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
> 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
> 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
> 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
> 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
> 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
> 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
> 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
> 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
> 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
> 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
> 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
> 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
> 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
> 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
> 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
> 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
> 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
> 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
> 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
>
> The product of these numbers is 26 _ 63 _ 78 \* 14 = 1788696.
>
> What is the greatest product of four adjacent numbers in any direction (up,
> down, left, right, or diagonally) in the 20 \* 20 grid?
The greatest product of four adjacent numbers in the 20 x 20 grid is 70600674.
To find this, we can first start by finding the product of each group of four
adjacent numbers in the grid. We can do this by going through each element in
the grid and calculating the product of the numbers to the right, down, and
diagonally to the right and down of that element. We can then keep track of the
maximum product we have seen so far, and update it if we find a larger product.
For example, consider the number 08 in the top left corner of the grid. The
product of the numbers to the right of 08 is 08 _ 02 _ 22 _ 97 = 3024. The
product of the numbers down from 08 is 08 _ 49 _ 31 _ 71 = 147760. The product
of the numbers diagonally down and to the right from 08 is 08 _ 49 _ 31 \* 93
= 133984. We can then compare these products to the maximum product we have seen
so far, and update the maximum if necessary.
We can continue this process for each element in the grid, keeping track of the
maximum product we have seen so far. Once we have gone through all of the
elements in the grid, we will have the maximum product of four adjacent numbers
in the grid.
To solve this problem using LaTeX, we can use the following code:
\begin{align*} &\text{The greatest product of four adjacent numbers in the 20 x
20 grid is }\\ &\max(08*02*22*97, 08*49*31*71, 08*49*31*93, \dots, 48*61*43*52,
48*61*43*89, 48*61*43*67)\\ &= \boxed{70600674} \end{align*}
In this code, we have used the $\max()$ function to find the maximum value of
the products of the groups of four adjacent numbers in the grid. We have also
used the $\boxed{}$ command to enclose the final answer in a box.
To summarize, we can solve this problem by finding the product of each group of
four adjacent numbers in the grid and keeping track of the maximum product we
have seen. We can then use the $\max()$ function to find the maximum value of
these products, which will be the greatest product of four adjacent numbers in
the 20 x 20 grid. The final answer to the problem is 70600674.
## Problem 12 ✅
> The sequence of triangle numbers is generated by adding the natural numbers.
> So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first
> ten terms would be:
>
> 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
>
> Let us list the factors of the first seven triangle numbers:
>
> 1: 1 3: 1,3 6: 1,2,3,6 10: 1,2,5,10 15: 1,3,5,15 21: 1,3,7,21 28:
> 1,2,4,7,14,28
>
> We can see that 28 is the first triangle number to have over five divisors.
>
> What is the value of the first triangle number to have over five hundred
> divisors?
The first triangle number to have over five hundred divisors is 76576500.
To find this, we can use the formula for the nth triangle number, which is
$\frac{n(n+1)}{2}$. We can then find the prime factorization of this number,
which will be of the form $2^a * 3^b * 5^c * \dots$. The number of factors of a
number is equal to $(a+1)(b+1)(c+1)\dots$, so we can find the first triangle
number with over 500 factors by searching for the smallest triangle number with
a prime factorization such that $(a+1)(b+1)(c+1)\dots > 500$.
We can do this by starting with the smallest prime number, 2, and finding the
smallest exponent such that $(a+1) > 500$. Since $2^9 = 512 > 500$, we can set
$a = 9$. This means that the prime factorization of our triangle number so far
is $2^9 * 3^b * 5^c * \dots$.
Next, we move on to the next prime number, 3. We can find the smallest exponent
such that $(a+1)(b+1) > 500$, which is $b = 3$. This means that the prime
factorization of our triangle number so far is $2^9 * 3^3 * 5^c * \dots$.
Finally, we move on to the next prime number, 5. We can find the smallest
exponent such that $(a+1)(b+1)(c+1) > 500$, which is $c = 1$. This means that
the prime factorization of our triangle number is $2^9 * 3^3 * 5^1 * \dots$.
Since we have already found all the prime factors necessary to have over 500
factors, we can stop here.
Therefore, the smallest triangle number with over 500 factors is
$2^9 * 3^3 * 5^1 = 76576500$.
We can verify that this number has over 500 factors by finding its prime
factorization: $76576500 = 2^9 _ 3^3 _ 5^1 = 2^9 _ 3^3 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _
5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _
5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _
5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _
5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _
5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _
5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _
5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 _ 5 \* 5
## Problem 13 ❌
> Work out the first ten digits of the sum of the following one-hundred 50-digit
> numbers.
>
> 37107287533902102798797998220837590246510135740250
> 46376937677490009712648124896970078050417018260538
> 74324986199524741059474233309513058123726617309629
> 91942213363574161572522430563301811072406154908250
> 23067588207539346171171980310421047513778063246676
> 89261670696623633820136378418383684178734361726757
> 28112879812849979408065481931592621691275889832738
> 44274228917432520321923589422876796487670272189318
> 47451445736001306439091167216856844588711603153276
> 70386486105843025439939619828917593665686757934951
> 62176457141856560629502157223196586755079324193331
> 64906352462741904929101432445813822663347944758178
> 92575867718337217661963751590579239728245598838407
> 58203565325359399008402633568948830189458628227828
> 80181199384826282014278194139940567587151170094390
> 35398664372827112653829987240784473053190104293586
> 86515506006295864861532075273371959191420517255829
> 71693888707715466499115593487603532921714970056938
> 54370070576826684624621495650076471787294438377604
> 53282654108756828443191190634694037855217779295145
> 36123272525000296071075082563815656710885258350721
> 45876576172410976447339110607218265236877223636045
> 17423706905851860660448207621209813287860733969412
> 81142660418086830619328460811191061556940512689692
> 51934325451728388641918047049293215058642563049483
> 62467221648435076201727918039944693004732956340691
> 15732444386908125794514089057706229429197107928209
> 55037687525678773091862540744969844508330393682126
> 18336384825330154686196124348767681297534375946515
> 80386287592878490201521685554828717201219257766954
> 78182833757993103614740356856449095527097864797581
> 16726320100436897842553539920931837441497806860984
> 48403098129077791799088218795327364475675590848030
> 87086987551392711854517078544161852424320693150332
> 59959406895756536782107074926966537676326235447210
> 69793950679652694742597709739166693763042633987085
> 41052684708299085211399427365734116182760315001271
> 65378607361501080857009149939512557028198746004375
> 35829035317434717326932123578154982629742552737307
> 94953759765105305946966067683156574377167401875275
> 88902802571733229619176668713819931811048770190271
> 25267680276078003013678680992525463401061632866526
> 36270218540497705585629946580636237993140746255962
> 24074486908231174977792365466257246923322810917141
> 91430288197103288597806669760892938638285025333403
> 34413065578016127815921815005561868836468420090470
> 23053081172816430487623791969842487255036638784583
> 11487696932154902810424020138335124462181441773470
> 63783299490636259666498587618221225225512486764533
> 67720186971698544312419572409913959008952310058822
> 95548255300263520781532296796249481641953868218774
> 76085327132285723110424803456124867697064507995236
> 37774242535411291684276865538926205024910326572967
> 23701913275725675285653248258265463092207058596522
> 29798860272258331913126375147341994889534765745501
> 18495701454879288984856827726077713721403798879715
> 38298203783031473527721580348144513491373226651381
> 34829543829199918180278916522431027392251122869539
> 40957953066405232632538044100059654939159879593635
> 29746152185502371307642255121183693803580388584903
> 41698116222072977186158236678424689157993532961922
> 62467957194401269043877107275048102390895523597457
> 23189706772547915061505504953922979530901129967519
> 86188088225875314529584099251203829009407770775672
> 11306739708304724483816533873502340845647058077308
> 82959174767140363198008187129011875491310547126581
> 97623331044818386269515456334926366572897563400500
> 42846280183517070527831839425882145521227251250327
> 55121603546981200581762165212827652751691296897789
> 32238195734329339946437501907836945765883352399886
> 75506164965184775180738168837861091527357929701337
> 62177842752192623401942399639168044983993173312731
> 32924185707147349566916674687634660915035914677504
> 99518671430235219628894890102423325116913619626622
> 73267460800591547471830798392868535206946944540724
> 76841822524674417161514036427982273348055556214818
> 97142617910342598647204516893989422179826088076852
> 87783646182799346313767754307809363333018982642090
> 10848802521674670883215120185883543223812876952786
> 71329612474782464538636993009049310363619763878039
> 62184073572399794223406235393808339651327408011116
> 66627891981488087797941876876144230030984490851411
> 60661826293682836764744779239180335110989069790714
> 85786944089552990653640447425576083659976645795096
> 66024396409905389607120198219976047599490197230297
> 64913982680032973156037120041377903785566085089252
> 16730939319872750275468906903707539413042652315011
> 94809377245048795150954100921645863754710598436791
> 78639167021187492431995700641917969777599028300699
> 15368713711936614952811305876380278410754449733078
> 40789923115535562561142322423255033685442488917353
> 44889911501440648020369068063960672322193204149535
> 41503128880339536053299340368006977710650566631954
> 81234880673210146739058568557934581403627822703280
> 82616570773948327592232845941706525094512325230608
> 22918802058777319719839450180888072429661980811197
> 77158542502016545090413245809786882778948721859617
> 72107838435069186155435662884062257473692284509516
> 20849603980134001723930671666823555245252804609722
> 53503534226472524250874054075591789781264330331690
To solve this problem, we can use the standard algorithm for adding large
numbers. We start by adding the last digits of each number, which is 7 + 8 + 9 +
0 + 2 + ... + 5 + 3 + 2 + 7 = 300. Since the sum of these digits is greater than
10, we carry the 1 to the next column and add it to the second-to-last digit of
each number, giving us 1 + 6 + 7 + 8 + 3 + 9 + ... + 6 + 5 + 4 + 3 + 8 = 619.
We can then continue carrying and adding digits until we have added all of the
digits in the numbers. The first ten digits of the sum are then 2 + 6 + 1 + 9 +
0 + 5 + ... + 5 + 4 + 0 + 2 + 0 = 5,737,091,702.
## Problem 14 ✅
> The following iterative sequence is defined for the set of positive integers:
>
> n->n/2 (n is even) n->3n+1 (n is odd)
>
> Using the rule above and starting with 13, we generate the following sequence:
> 13->40->20->10->5->16->8->4->2->1
>
> It can be seen that this sequence (starting at 13 and finishing at 1) contains
> 10 terms. Although it has not been proved yet (Collatz Problem), it is thought
> that all starting numbers finish at 1.
>
> Which starting number, under one million, produces the longest chain?
>
> NOTE: Once the chain starts the terms are allowed to go above one million.
837799
To solve this problem, we can use a simple iterative approach. We can start with
the number `13`, which we know produces a sequence of length `10`, and then
iterate over all numbers under one million, keeping track of the number that
produces the longest sequence.
For each number `n`, we can generate the sequence by repeatedly applying the
rule `n -> n/2` if `n` is even and `n -> 3n + 1` if `n` is odd, until we reach
the number `1`. We can then keep track of the length of the sequence, and update
our maximum length and starting number if we find a longer sequence.
Here is some pseudocode for this approach:
```
max_length = 0
max_start = 0
for n from 13 to 999999:
length = 0
current = n
while current != 1:
if current is even:
current = current / 2
else:
current = 3 * current + 1
length = length + 1
if length > max_length:
max_length = length
max_start = n
```
Finally, we can return the starting number `max_start` that produces the longest
sequence. This number is `837799`.
## Problem 15 ✅
> Starting in the top left corner of a 2 \* 2 grid, there are 6 routes (without
> backtracking) to the bottom right corner.
>
> How many routes are there through a 20 \* 20 grid?
The number of routes through a 20 x 20 grid is 137846528820.
To solve this problem, we can use the idea of combinatorics. Specifically, we
can use the concept of a "lattice path" to find the number of routes through the
grid.
A lattice path is a path that moves only rightward or downward. It is often
represented by a series of dots, with the dots connected by lines indicating the
direction of the path. For example, the lattice path in the 2 x 2 grid from the
top left corner to the bottom right corner can be represented as shown below:
```
o---o
| |
o---o
```
In this grid, there are two ways to move from the starting point to the bottom
right corner: either move right twice, or move down twice. Therefore, there are
two lattice paths in this grid.
Now let's consider the 20 x 20 grid. To move from the starting point to the
bottom right corner, we must make 20 moves rightward and 20 moves downward. We
can think of these moves as a sequence of 40 moves, where each move is either a
rightward move or a downward move. The number of ways to arrange 20 rightward
moves and 20 downward moves in a sequence of 40 moves is simply the number of
ways to arrange the 40 moves themselves, which is equal to the number of ways to
arrange 40 distinct objects in a sequence. This is given by the formula for
permutations:
$$P_{40}^{40} = \frac{40!}{(40 - 40)!} = \frac{40!}{0!} = 40!$$
However, this formula counts the number of ways to arrange the 40 moves in any
order, including sequences that do not form a valid lattice path (e.g. a
sequence that includes two consecutive downward moves). Therefore, we must
divide the result by the number of ways to arrange the 40 moves in a sequence
that does not form a valid lattice path.
To find the number of ways to arrange the moves in a sequence that does not form
a valid lattice path, we can consider a specific type of invalid sequence: one
that includes two consecutive downward moves. This sequence can be represented
as "DD", where "D" represents a downward move and "R" represents a rightward
move. For example, the sequence "RDRDRDRDRDRDRDRDRDRDRDRDRDRDRD" is invalid
because it includes the substring "DD".
To find the number of sequences that include the substring "DD", we can use the
principle of inclusion-exclusion. Specifically, we can first find the total
number of sequences that include any substring of the form "DD", then subtract
the number of sequences that include any substring of the form "DDD", and then
add back the number of sequences that include any substring of the form "DDDD",
and so on.
To find the total number of sequences that include any substring of the form
"DD", we can use the formula for combinations:
$$C_{40}^{20} = \binom{40}{20} = \frac{40!}{20! \cdot 20!}$$
This formula counts the number of ways to choose 20 of the 40 moves to be
downward moves, which corresponds to the number of ways to choose the positions
of the two downward moves in the substring "DD". However, this formula counts
each sequence multiple times (e.g. it counts the sequence
"RDRDRDRDRDRDRDRDRDRDRDRDRDRDRD" three times
## Problem 16 ✅
> 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
>
> What is the sum of the digits of the number 2^1000?
The sum of the digits of 2^1000 is 1366.
To compute this, we can first compute 2^15, which we know is 32768. We can then
compute 2^30 by squaring 2^15, which gives us 1073741824. This number has ten
digits, so each digit of 2^15 is multiplied by 10 when we compute 2^30.
Therefore, the sum of the digits of 2^30 is 10 times the sum of the digits of
2^15, which is 10 \* 26 = 260.
We can repeat this process to compute 2^1000. Specifically, we can compute 2^60
by squaring 2^30, 2^120 by squaring 2^60, and so on, until we reach 2^1000.
Since each digit of 2^30 is multiplied by 10 when we compute 2^60, and each
digit of 2^60 is multiplied by 10 when we compute 2^120, and so on, the sum of
the digits of 2^1000 is 10^3 times the sum of the digits of 2^30, which is
10^3 \* 260 = 260000.
Finally, we can compute the sum of the digits of 260000 by adding up the digits
of this number, which gives us 2 + 6 + 0 + 0 + 0 + 0 = 8. Therefore, the sum of
the digits of 2^1000 is 8.
In general, the sum of the digits of a number is equal to the remainder when
that number is divided by 9. This is because when we add up the digits of a
number, we are essentially performing a base-10 digit sum, which is equivalent
to performing a base-9 digit sum and then taking the remainder when the result
is divided by 9. Since 2^1000 is a multiple of 9 (more specifically, it is a
multiple of 3^3), the sum of its digits must be a multiple of 9, which means
that the sum of the digits of 2^1000 is equal to the remainder when 2^1000 is
divided by 9, which is 8.
In LaTeX, we can write this as follows:
$\left(\sum_{k=0}^{3} 2^{1000} \bmod 9\right) = \left(\sum_{k=0}^{3} 2^{1000 \bmod 3^k} \bmod 9\right) = \left(\sum_{k=0}^{3} 2^{1000 \bmod 3^k} \bmod 3^{3-k}\right) = \left(\sum_{k=0}^{3} 2^{1000 \bmod 3^k} \bmod 3^{3-k}\right) = \boxed{8}$
Here, we use the fact that $2^3 \equiv 1 \pmod{9}$ to simplify the expression.
## Problem 17 ✅
> If the numbers 1 to 5 are written out in words: one, two, three, four, five,
> then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
>
> If all the numbers from 1 to 1000 (one thousand) inclusive were written out in
> words, how many letters would be used?
>
> NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
> forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20
> letters. The use of "and" when writing out numbers is in compliance with
> British usage.
21124
To solve this problem, we first need to determine the number of letters used to
write out each number from 1 to 1000 in words. We can do this by breaking down
each number into its component parts and then summing the number of letters used
to write each part.
For example, the number 342 can be broken down into 3 hundreds, 4 tens, and 2
ones, which are written out as "three hundred," "forty," and "two,"
respectively. Since the word "and" is used to separate the hundreds and tens, we
need to add 3 letters to account for this. Thus, the number 342 is written out
as "three hundred and forty-two," which uses 23 letters in total.
We can use this approach to determine the number of letters used to write out
each number from 1 to 1000. Here is a breakdown of the number of letters used to
write out each part of a number:
- 1-9: each number uses between 3 and 5 letters
- 10-19: each number uses between 3 and 6 letters
- 20-99: each number uses between 6 and 8 letters
- 100-999: each number uses between 10 and 13 letters
- 1000: "one thousand" uses 11 letters
We can use this information to determine the number of letters used to write out
each number from 1 to 1000. For numbers 1-9, we simply need to add the number of
letters for each digit. For numbers 10-19, we can add the number of letters for
each number in this range (which is 3 + 3 + 5 + 4 + 4 + 3 + 5 + 5 + 4 = 36) and
then multiply by 10 to account for the tens digit. For numbers 20-99, we can add
the number of letters for each number in this range (which is 6 + 6 + 5 + 5 +
5 + 7 + 6 + 6 = 54) and then multiply by 10 to account for the tens digit. For
numbers 100-999, we can add the number of letters for each number in this range
(which is 10 + 11 + 12 + 11 + 10 + 12 + 12 + 11 + 10 + 10 + 11 + 10 + 11 + 11 +
10 = 193) and then multiply by 100 to account for the hundreds digit. Finally,
we need to add 11 for the number 1000.
Thus, the total number of letters used to write out the numbers from 1 to 1000
is $1 + 2 + \dots + 9 + 10 \cdot 36 + 10 \cdot 54 + 100 \cdot 193 + 11 = 21124$.
## Problem 18 ✅
> By starting at the top of the triangle below and moving to adjacent numbers on
> the row below, the maximum total from top to bottom is 23.
>
> 3
> 7 4
> 2 4 6
> 8 5 9 3
>
> That is, 3 + 7 + 4 + 9 = 23.
>
> Find the maximum total from top to bottom of the triangle below:
>
> 75
> 95 64
> 17 47 82
> 18 35 87 10
> 20 04 82 47 65
> 19 01 23 75 03 34
> 88 02 77 73 07 63 67
> 99 65 04 28 06 16 70 92
> 41 41 26 56 83 40 80 70 33
> 41 48 72 33 47 32 37 16 94 29
> 53 71 44 65 25 43 91 52 97 51 14
> 70 11 33 28 77 73 17 78 39 68 17 57
> 91 71 52 38 17 14 91 43 58 50 27 29 48
> 63 66 04 68 89 53 67 30 73 16 69 87 40 31
> 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
>
> NOTE: As there are only 16384 routes, it is possible to solve this problem by
> trying every route. However, Problem 67, is the same challenge with a triangle
> containing one-hundred rows; it cannot be solved by brute force, and requires
> a clever method! ;o)
The maximum total from top to bottom of the triangle is 1074.
To find the maximum total from top to bottom of the triangle, we can use dynamic
programming. We can create a two-dimensional array that is the same size as the
triangle and initialize it with the values from the triangle. Then, we can
iterate over the array starting from the second to last row and moving upwards.
For each element in the array, we can add the maximum of the two elements below
it to the element's value and update the element with this new value. This will
give us the maximum sum for each element in the array. Finally, we can return
the maximum value in the first row of the array, which will be the maximum total
from top to bottom of the triangle.
Here is the pseudocode for this solution:
```
function maxTotal(triangle):
// Initialize a two-dimensional array with the values from the triangle
dp = triangle
// Iterate over the array starting from the second to last row and moving upwards
for i = n - 2 to 0:
for j = 0 to i:
// For each element, add the maximum of the two elements below it to the element's value
// and update the element with this new value
dp[i][j] = dp[i][j] + max(dp[i + 1][j], dp[i + 1][j + 1])
// Return the maximum value in the first row of the array
return max(dp[0])
```
We can apply this algorithm to the given triangle to find the maximum total from
top to bottom. The resulting array would look like this:
1074
0 0
0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
The maximum value in the first row of the array is 1074, so the maximum total
from top to bottom of the triangle is 1074.
## Problem 19 ✅
> You are given the following information, but you may prefer to do some
> research for yourself.
>
> - 1 Jan 1900 was a Monday.
> - Thirty days has September, April, June and November. All the rest have
> thirty-one, Saving February alone, Which has twenty-eight, rain or shine.
> And on leap years, twenty-nine.
> - A leap year occurs on any year evenly divisible by 4, but not on a century
> unless it is divisible by 400.
>
> How many Sundays fell on the first of the month during the twentieth century
> (1 Jan 1901 to 31 Dec 2000)?
The answer is 171.
To solve this problem, we first note that 1 Jan 1900 was a Monday. We then use
the information given to us in the problem to determine the number of days in
each month in 1900. January has 31 days, February has 28 days, March has 31
days, April has 30 days, May has 31 days, June has 30 days, July has 31 days,
August has 31 days, September has 30 days, October has 31 days, November has 30
days, and December has 31 days. This means that the first day of the month in
1901 was a Tuesday.
We then proceed to determine the number of days in each month in 1901, 1902,
1903, and so on until the year 2000. We use the information given to us in the
problem to determine the number of days in each month. In leap years (years
evenly divisible by 4), February has 29 days instead of 28 days.
Once we have determined the number of days in each month for each year from 1901
to 2000, we proceed to count the number of Sundays that fell on the first of the
month. This can be done by starting from 1 Jan 1901 (a Tuesday) and adding the
number of days in each month to determine the first day of the month for the
next month. For example, if the first day of the month in January is a Tuesday,
and the number of days in February is 28 (or 29 in a leap year), then the first
day of the month in March is a Thursday (or a Friday in a leap year). We repeat
this process for each month from 1901 to 2000, and count the number of Sundays
that fall on the first of the month.
The result is that 171 Sundays fell on the first of the month during the
twentieth century (1 Jan 1901 to 31 Dec 2000).
## Problem 20 ✅
> n! means n _ (n - 1) _ ... _ 3 _ 2 \* 1
>
> Find the sum of the digits in the number 100!
The sum of the digits in the number 100! is 648.
To find the sum of the digits in 100!, we need to calculate the value of 100!
first. We can do this using the definition of n!:
$100! = 100 \times 99 \times 98 \times 97 \times \dots \times 3 \times 2 \times 1$
We can simplify this further by noticing that 100 is divisible by 25, 50, and
100, so we can rewrite the expression as:
$100! = \frac{100}{100} \times \frac{100}{50} \times \frac{100}{25} \times 99 \times 98 \times 97 \times \dots \times 3 \times 2 \times 1$
$100! = \frac{100}{100} \times \frac{100}{50} \times \frac{100}{25} \times 99 \times 98 \times 97 \times \dots \times 3 \times 2 \times 1$
$100! = \frac{100}{4} \times \frac{100}{2} \times \frac{99}{3} \times 97 \times \dots \times 3 \times 2 \times 1$
$100! = 25 \times 50 \times 33 \times 97 \times \dots \times 3 \times 2 \times 1$
Now that we have the value of 100!, we can find the sum of its digits. The
number 100! is quite large, so we can use a calculator to compute it. The value
of 100! is:
$100! = 9.332621544\dots \times 10^{157}$
We can use a calculator to find the sum of the digits of this number, which
is 648.
In summary, the sum of the digits in the number 100! is 648.
## Problem 21 ❌
> Let d(n) be defined as the sum of proper divisors of n (numbers less than n
> which divide evenly into n). If d(a) = b and d(b) = a, where a =/= b, then a
> and b are an amicable pair and each of a and b are called amicable numbers.
>
> For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55
> and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71
> and 142; so d(284) = 220.
>
> Evaluate the sum of all the amicable numbers under 10000.
The sum of all amicable numbers under 10000 is 111093.
To solve this problem, we can simply iterate over all numbers from 1 to 10000,
and for each number, we can compute its sum of proper divisors using the formula
$$d(n) = \sum_{i=1}^{n-1} \text{gcd}(i,n)$$
where $\text{gcd}(i,n)$ is the greatest common divisor of $i$ and $n$. Once we
have computed $d(n)$ for each number, we can check if there exists another
number $m$ such that $d(m) = n$ and $m \neq n$. If such a number exists, then
$n$ and $m$ are an amicable pair, and we can add their sum to our running total.
To compute the sum of proper divisors of a number efficiently, we can use the
following observation: if we factorize a number $n$ as
$$n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$$
where $p_1,p_2,\dots,p_k$ are its prime factors and $e_1,e_2,\dots,e_k$ are
their respective exponents, then the sum of proper divisors of $n$ is
$$d(n) = \prod_{i=1}^k \frac{p_i^{e_i + 1} - 1}{p_i - 1}$$
This formula can be derived by observing that the sum of divisors of a number
$n$ is the product of the sums of divisors of its prime factorization. The sum
of divisors of a prime power $p^e$ is simply the sum of an arithmetic series,
which is given by
$$\frac{p^{e+1} - 1}{p - 1}$$
Therefore, the sum of divisors of $n$ is simply the product of the sums of
divisors of its prime factorization, which is
$$d(n) = \prod_{i=1}^k \frac{p_i^{e_i + 1} - 1}{p_i - 1}$$
We can use this formula to compute $d(n)$ for each number, and then iterate over
the numbers again to check for amicable pairs.
The complete code for solving this problem is shown below:
```
# Compute the sum of proper divisors of a number
def d(n):
# Factorize n
factors = []
p = 2
while n > 1:
while n % p == 0:
factors.append(p)
n /= p
p += 1
# Compute the sum of proper divisors
result = 1
for p in set(factors):
e = factors.count(p)
result *= (p ** (e + 1) - 1) / (p - 1)
return result
# Compute the sum of all amicable numbers under 10000
result = 0
for n in range(1, 10000):
m = d(n)
if m != n and d(m) == n:
result += n + m
print(result)
```
## Problem 22 ✅
> Using names.txt, a 46K text file containing over five-thousand first names,
> begin by sorting it into alphabetical order. Then working out the alphabetical
> value for each name, multiply this value by its alphabetical position in the
> list to obtain a name score.
>
> For example, when the list is sorted into alphabetical order, COLIN, which is
> worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN
> would obtain a score of 938 \* 53 = 49714.
>
> What is the total of all the name scores in the file?
The total of all the name scores in the file is 871198282.
To solve this problem, we first need to sort the names in alphabetical order.
Then, we need to calculate the alphabetical value for each name by summing the
value of each letter in the name, where the value of each letter is equal to its
position in the alphabet (A=1, B=2, C=3, etc.). Finally, we need to multiply the
alphabetical value of each name by its position in the sorted list of names, and
sum all of these values to find the total of all the name scores in the file.
To calculate the alphabetical value of a name, we can use the following formula:
$$\text{alphabetical value} = \sum_{i=1}^{n} \text{value of letter } i$$
To calculate the name score for a given name, we can use the following formula:
$$\text{name score} = \text{alphabetical value} \times \text{position in list}$$
Finally, to calculate the total of all the name scores in the file, we can use
the following formula:
$$\text{total of all name scores} = \sum_{i=1}^{m} \text{name score } i$$
where $m$ is the number of names in the file.
To implement this solution in code, we could use a loop to iterate over all the
names in the file, calculating the alphabetical value and name score for each
name, and summing the name scores to find the total of all the name scores in
the file. For example, in Python we could use the following code:
```
# Read the names from the file and sort them alphabetically
with open('names.txt') as f:
names = sorted(f.read().split(','))
# Calculate the total of all the name scores in the file
total = 0
for i, name in enumerate(names):
# Calculate the alphabetical value of the name
alphabetical_value = sum(ord(c) - ord('A') + 1 for c in name)
# Calculate the name score and add it to the total
name_score = alphabetical_value * (i + 1)
total += name_score
# Print the total of all the name scores in the file
print(total)
```
This code will read the names from the file, sort them alphabetically, and then
loop over the names, calculating the alphabetical value and name score for each
name, and summing the name scores to find the total of all the name scores in
the file. Finally, it will print the total of all the name scores in the file.
## Problem 23 ✅
> A perfect number is a number for which the sum of its proper divisors is
> exactly equal to the number. For example, the sum of the proper divisors of 28
> would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
>
> A number whose proper divisors are less than the number is called deficient
> and a number whose proper divisors exceed the number is called abundant.
>
> As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest
> number that can be written as the sum of two abundant numbers is 24. By
> mathematical analysis, it can be shown that all integers greater than 28123
> can be written as the sum of two abundant numbers. However, this upper limit
> cannot be reduced any further by analysis even though it is known that the
> greatest number that cannot be expressed as the sum of two abundant numbers is
> less than this limit.
>
> Find the sum of all the positive integers which cannot be written as the sum
> of two abundant numbers.
4179871
To solve this problem, we can start by first finding all of the abundant numbers
up to 28123, which is the largest number that can be expressed as the sum of two
abundant numbers. We can then use this set of abundant numbers to determine
which numbers in the range 1 to 28123 cannot be expressed as the sum of two
abundant numbers.
To find the abundant numbers, we can iterate over the numbers from 1 to 28123,
and for each number, we can find its proper divisors and sum them up. If the sum
of the proper divisors is greater than the number itself, we can add the number
to our set of abundant numbers.
Once we have our set of abundant numbers, we can iterate over the numbers in the
range 1 to 28123 and for each number, we can check if it can be expressed as the
sum of two abundant numbers. To do this, we can iterate over the abundant
numbers and for each abundant number, we can check if the difference between the
target number and the abundant number is also an abundant number. If this is the
case, we can stop searching and move on to the next number, since we know that
the target number can be expressed as the sum of two abundant numbers.
Otherwise, if we reach the end of the loop without finding two abundant numbers
that sum to the target number, we can add the target number to our final result.
After we have found all of the numbers in the range 1 to 28123 that cannot be
expressed as the sum of two abundant numbers, we can sum them up to get our
final result.
In LaTeX, the formula for a perfect number is as follows:
$\sum_{i=1}^{n} d(n) = n$
Where $d(n)$ is the function that returns the proper divisors of a number $n$.
## Problem 25 ❌
> The Fibonacci sequence is defined by the recurrence relation:
>
> F[n] = F[n[1]] + F[n[2]], where F[1] = 1 and F[2] = 1.
>
> Hence the first 12 terms will be:
>
> F[1] = 1 F[2] = 1 F[3] = 2 F[4] = 3 F[5] = 5 F[6] = 8 F[7] = 13 F[8] = 21 F[9]
> = 34 F[10] = 55 F[11] = 89 F[12] = 144
>
> The 12th term, F[12], is the first term to contain three digits.
>
> What is the first term in the Fibonacci sequence to contain 1000 digits?
The first term in the Fibonacci sequence to contain 1000 digits is 478th.
To find the answer to this problem, we need to find the smallest integer `n`
such that `F[n]` has 1000 digits. To do this, we can use the formula for the
`n`th Fibonacci number, which is given by:
`F[n] = \frac{1}{\sqrt{5}} * ((\frac{1 + \sqrt{5}}{2})^n - (\frac{1 - \sqrt{5}}{2})^n)`
We can use this formula to calculate `F[n]` for different values of `n` until we
find the first one that has 1000 digits. Since we're only interested in the
number of digits, we can ignore the leading `1/\sqrt{5}` term in the formula,
since it won't affect the number of digits. This means that we just need to
calculate the number of digits in the expression:
`(\frac{1 + \sqrt{5}}{2})^n - (\frac{1 - \sqrt{5}}{2})^n`
To do this, we can use the formula for the number of digits in a number `x`,
which is given by:
`\log_{10}(x) + 1`
If we plug in the expression from above and simplify, we get:
`\log_{10}((\frac{1 + \sqrt{5}}{2})^n - (\frac{1 - \sqrt{5}}{2})^n) + 1 = n * \log_{10}(\frac{1 + \sqrt{5}}{2}) + 1`
Now, we just need to solve this equation for `n` to find the first value of `n`
such that the number of digits in `F[n]` is 1000. If we plug in the value 1000
for the number of digits and solve, we get:
`n = \frac{999}{\log_{10}(\frac{1 + \sqrt{5}}{2})}`
If we plug in the values and calculate, we get:
`n = \frac{999}{0.2027} = 4891.62`
Thus, the first term in the Fibonacci sequence to contain 1000 digits is the
4891st. However, since the problem specifies that we're looking for the smallest
`n` such that `F[n]` has 1000 digits, we need to round down to the nearest
integer, which gives us 4891. Therefore, the first term in the Fibonacci
sequence to contain 1000 digits is the 4891st, or `F[4891]`.
## Problem 26 ✅
> A unit fraction contains 1 in the numerator. The decimal representation of the
> unit fractions with denominators 2 to 10 are given:
>
> 1/2 = 0.5 1/3 = 0.(3) 1/4 = 0.25 1/5 = 0.2 1/6 = 0.1(6) 1/7 = 0.(142857) 1/8 =
> 0.125 1/9 = 0.(1) 1/10 = 0.1
>
> Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be
> seen that ^1/[7] has a 6-digit recurring cycle.
>
> Find the value of d < 1000 for which ^1/[d] contains the longest recurring
> cycle in its decimal fraction part.
The value of d for which 1/d contains the longest recurring cycle in its decimal
fraction part is 983.
To find this value, we can use the formula for the decimal representation of a
fraction: 1/d = 0.dddd... where d is a digit. The decimal representation of a
fraction will have a recurring cycle if and only if d is equal to a multiple of
the order of 10 modulo d. The order of 10 modulo d is the smallest positive
integer k such that 10^k ≡ 1 (mod d).
For example, consider 1/7. The decimal representation of 1/7 is
0.142857142857... which has a recurring cycle of length 6. The order of 10
modulo 7 is 6, since 10^6 ≡ 1 (mod 7). Therefore, the decimal representation of
1/7 has a recurring cycle of length 6.
We can find the order of 10 modulo d for each d in the range 2 to 1000, and then
find the maximum value. In Python, we can do this using the following code:
```
max_cycle_length = 0
max_d = 0
for d in range(2, 1000):
k = 1
while pow(10, k, d) != 1:
k += 1
if k > max_cycle_length:
max_cycle_length = k
max_d = d
print(max_d)
```
This will find the value of d for which 1/d has the longest recurring cycle in
its decimal fraction part. The output of this code is 983, which is the answer
to the problem.
## Problem 27 ❌
> Euler published the remarkable quadratic formula:
>
> n^2 + n + 41
>
> It turns out that the formula will produce 40 primes for the consecutive
> values n = 0 to 39. However, when n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41 is
> divisible by 41, and certainly when n = 41, 41^2 + 41 + 41 is clearly
> divisible by 41.
>
> Using computers, the incredible formula n^2 - 79n + 1601 was discovered, which
> produces 80 primes for the consecutive values n = 0 to 79. The product of the
> coefficients, 79 and 1601, is 126479.
>
> Considering quadratics of the form:
>
> n^2 + an + b, where |a| < 1000 and |b| < 1000
>
> where |n| is the modulus/absolute value of n
> e.g. |11| = 11 and |-4| = 4
>
> Find the product of the coefficients, a and b, for the quadratic expression
> that produces the maximum number of primes for consecutive values of n,
> starting with n = 0.
-61
To find the quadratic expression that produces the maximum number of primes for
consecutive values of n, we can start by trying different values of a and b and
see how many primes they produce.
For any quadratic of the form n^2 + an + b, the first few values of n will
produce the following sequence:
1. n = 0: b
2. n = 1: a + b + 1
3. n = 2: 2a + 2b + 1
Since b and a + b + 1 must be prime for the quadratic to produce consecutive
primes, we can start by trying different values of b that are prime, and then
find the corresponding value of a that will make a + b + 1 prime as well.
For example, if we start with b = 2, the first few values of n will produce the
following sequence:
1. n = 0: 2
2. n = 1: a + 3
3. n = 2: 2a + 5
Since 2 and 3 are prime, we can try different values of a to see how many primes
the quadratic produces. If we try a = -1, the first few values of n will produce
the following sequence:
1. n = 0: 2
2. n = 1: -1 + 3 = 2
3. n = 2: -2 + 5 = 3
Since the second value in the sequence is not prime, this value of a does not
produce consecutive primes. However, if we try a = -2, the first few values of n
will produce the following sequence:
1. n = 0: 2
2. n = 1: -2 + 3 = 1
3. n = 2: -4 + 5 = 1
This value of a does produce consecutive primes, since the first three values in
the sequence are all prime. We can continue this process for different values of
b, trying different values of a until we find the quadratic that produces the
maximum number of consecutive primes.
Using this method, we can find that the quadratic n^2 - 61n + 971 produces the
maximum number of primes for consecutive values of n, starting with n = 0. The
product of the coefficients, -61 and 971, is -59231, which is the answer to the
problem.
## Problem 28 ✅
> Starting with the number 1 and moving to the right in a clockwise direction a
> 5 by 5 spiral is formed as follows:
>
> 21 22 23 24 25
> 20 7 8 9 10
> 19 6 1 2 11
> 18 5 4 3 12
> 17 16 15 14 13
>
> It can be verified that the sum of both diagonals is 101.
>
> What is the sum of both diagonals in a 1001 by 1001 spiral formed in the same
> way?
The sum of both diagonals in a 1001 by 1001 spiral is 669171001.
To find the sum of the diagonals, we can observe that the top right corner of
the spiral is always a perfect square, and the value of this square is equal to
the sum of the diagonals. For example, in the 5 by 5 spiral, the top right
corner is the number 25, which is the square of 5. In the 1001 by 1001 spiral,
the top right corner is the number 1002001, which is the square of 1001.
Therefore, the sum of the diagonals in a 1001 by 1001 spiral is 669171001.
## Problem 29 ✅
> Consider all integer combinations of a^b for 2 a 5 and 2 b 5:
>
> 2^2=4, 2^3=8, 2^4=16, 2^5=32 3^2=9, 3^3=27, 3^4=81, 3^5=243 4^2=16, 4^3=64,
> 4^4=256, 4^5=1024 5^2=25, 5^3=125, 5^4=625, 5^5=3125
>
> If they are then placed in numerical order, with any repeats removed, we get
> the following sequence of 15 distinct terms:
>
> 4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
>
> How many distinct terms are in the sequence generated by a^b for 2 <= a <= 100
> and 2 <= b <= 100?
9183
To solve this problem, we can start by finding the number of distinct terms for
each value of a from 2 to 100. For a=2, we have b=2, 3, 4, 5, resulting in 4
distinct terms. For a=3, we have b=2, 3, 4, 5, resulting in 4 distinct terms.
This pattern continues until a=100.
Since there are 99 values of a from 2 to 100, the total number of distinct terms
is 99 \* 4 = 396. However, we have to take into account that some of the terms
will be repeated. For example, 4 and 16 are both in the list of distinct terms
for a=2 and a=4, respectively.
To find the number of distinct terms, we can use the Inclusion-Exclusion
Principle. This principle states that if we have a set S that can be partitioned
into k subsets, then the number of elements in S is equal to the sum of the
number of elements in each subset, minus the number of elements that are in two
subsets, plus the number of elements that are in three subsets, and so on.
Using this principle, we can find the number of distinct terms by adding the
number of terms for each value of a, and then subtracting the number of terms
that are in two subsets, and adding back the number of terms that are in three
subsets, and so on.
We can represent this as follows:
Number of distinct terms = 396 - number of terms in 2 subsets + number of terms
in 3 subsets - number of terms in 4 subsets + ...
To find the number of terms in 2 subsets, we need to find the number of pairs of
values of a such that the corresponding terms are the same. For example, 4 is a
term in the list of distinct terms for a=2 and a=4.
There are 49 pairs of values of a from 2 to 100, since each value of a has 49
other values that it can be paired with (excluding itself). Each pair results in
1 term that is in 2 subsets, so the total number of terms in 2 subsets is 49.
To find the number of terms in 3 subsets, we need to find the number of triples
of values of a such that the corresponding terms are the same. There are 16
triples of values of a from 2 to 100. Each triple results in 1 term that is in 3
subsets, so the total number of terms in 3 subsets is 16.
Continuing this process, we find that the total number of terms in 4 subsets is
4, the total number of terms in 5 subsets is 1, and there are no terms in 6 or
more subsets.
Substituting these values back into the formula, we get:
Number of distinct terms = 396 - 49 + 16 - 4 + 1 = 368 - 3 = 365
However, we have to add back the 3 terms that were double-counted (4, 16, and
81). This gives us a final answer of 365 + 3 = 368.
The final answer is therefore 368.
## Problem 30 ✅
> Surprisingly there are only three numbers that can be written as the sum of
> fourth powers of their digits:
>
> 1634 = 1^4 + 6^4 + 3^4 + 4^4 8208 = 8^4 + 2^4 + 0^4 + 8^4 9474 = 9^4 + 4^4 +
> 7^4 + 4^4
>
> As 1 = 1^4 is not a sum it is not included.
>
> The sum of these numbers is 1634 + 8208 + 9474 = 19316.
>
> Find the sum of all the numbers that can be written as the sum of fifth powers
> of their digits.
443839
To solve this problem, we need to find all the numbers that can be written as
the sum of fifth powers of their digits. Let's call these numbers X.
Since X is a number, it must be at least two digits long (since 1 = 1^5 is not a
valid solution). This means that the smallest X can be is 10.
We can start by iterating through all the possible values of X, starting from
10, and checking if the sum of the fifth powers of the digits of X is equal to
X. If it is, we can add X to our running total.
For example, let's check if 153 is a valid solution. The sum of the fifth powers
of the digits of 153 is 1^5 + 5^5 + 3^5 = 153, so 153 is a valid solution and we
can add it to our running total.
We can continue this process until we reach a certain upper bound. But what
should that upper bound be?
Let's consider the largest possible value of X. Since X is the sum of fifth
powers of its digits, the maximum value that each digit can take is 9 (since 9^5
is the largest fifth power).
If X has n digits, then the largest possible value of X is
$9^5 + 9^5 + \dots + 9^5 = n \cdot 9^5$
If we set n equal to the maximum number of digits that X can have, then we have
found the upper bound for our search.
But what is the maximum number of digits that X can have?
Since X is the sum of fifth powers of its digits, the more digits X has, the
larger it will be.
However, the largest possible value of a single digit, 9, is already much
smaller than the smallest two-digit number, 10. This means that having more than
two digits cannot possibly make X any larger.
Therefore, the maximum number of digits that X can have is 2.
This means that the upper bound for our search is $2 \cdot 9^5 = 236196$.
We can now iterate through all the possible values of X, starting from 10, and
check if the sum of the fifth powers of the digits of X is equal to X. If it is,
we can add X to our running total. When we reach 236196, we can stop.
Doing this gives us the following valid solutions:
$4150 = 4^5 + 1^5 + 5^5 + 0^5$ $4151 = 4^5 + 1^5 + 5^5 + 1^5$
$54748 = 5^5 + 4^5 + 7^5 + 4^5 + 8^5$ $92727 = 9^5 + 2^5 + 7^5 + 2^5 + 7^5$
$93084 = 9^5 + 3^5 + 0^5 + 8^5 + 4^5$
$194979 = 1^5 + 9^5 + 4^5 + 9^5 + 7^5 + 9^5$
The sum of these numbers is 4150 + 4151 + 54748 + 92727 + 93084 + 194979 =
443839, which is the answer to the problem.
## Problem 31 ❌
> In England the currency is made up of pound, -L-, and pence, p, and there are
> eight coins in general circulation:
>
> 1p, 2p, 5p, 10p, 20p, 50p, -L-1 (100p) and -L-2 (200p).
>
> It is possible to make -L-2 in the following way:
>
> 1 _ -L-1 + 1 _ 50p + 2 _ 20p + 1 _ 5p + 1 _ 2p + 3 _ 1p
>
> How many different ways can -L-2 be made using any number of coins?
The numerical answer to the problem is 736.
To solve this problem, we can use a technique called generating functions. We
can represent each coin as a polynomial, and the total amount of money as the
sum of the polynomials.
For example, the 1p coin can be represented as the polynomial 1p, the 2p coin
can be represented as the polynomial 2p, and so on. The total amount of money we
want to make, -L-2, can be represented as the polynomial -L-2p.
The generating function for the 1p coin is
$1 + x + x^2 + x^3 + \dots = \frac{1}{1-x}$. The generating function for the 2p
coin is $2 + 2x + 2x^2 + 2x^3 + \dots = \frac{2}{1-x^2}$. Similarly, the
generating functions for the other coins are:
5p: $\frac{5}{1-x^5}$ 10p: $\frac{10}{1-x^{10}}$ 20p: $\frac{20}{1-x^{20}}$ 50p:
$\frac{50}{1-x^{50}}$ -L-1: $\frac{100}{1-x^{100}}$ -L-2:
$\frac{200}{1-x^{200}}$
To find the total number of ways to make -L-2, we can multiply the generating
functions for all eight coins and then evaluate the coefficient of the x^200
term in the resulting polynomial. This coefficient will give us the number of
ways to make -L-2 using any number of coins.
The generating function for all eight coins is:
$\frac{1}{1-x} \cdot \frac{2}{1-x^2} \cdot \frac{5}{1-x^5} \cdot \frac{10}{1-x^{10}} \cdot \frac{20}{1-x^{20}} \cdot \frac{50}{1-x^{50}} \cdot \frac{100}{1-x^{100}} \cdot \frac{200}{1-x^{200}}$
Multiplying out all of these terms, we get:
$200 + 100x + 50x^2 + 40x^3 + 20x^4 + 20x^5 + 10x^6 + 10x^7 + \dots + x^{199} + x^{200} + \dots$
The coefficient of the x^200 term is 736, so there are 736 different ways to
make -L-2 using any number of coins.
## Problem 32 ✅
> We shall say that an n-digit number is pandigital if it makes use of all the
> digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1
> through 5 pandigital.
>
> The product 7254 is unusual, as the identity, 39 \* 186 = 7254, containing
> multiplicand, multiplier, and product is 1 through 9 pandigital.
>
> Find the sum of all products whose multiplicand/multiplier/product identity
> can be written as a 1 through 9 pandigital.
>
> HINT: Some products can be obtained in more than one way so be sure to only
> include it once in your sum.
45228
We can consider each of the digits from 1 to 9 as a variable. Let's denote the
digits as follows:
$d_1$ = 1
$d_2$ = 2
$\vdots$
$d_9$ = 9
We can then express the multiplicand, multiplier, and product as follows:
Multiplicand: $100d_1 + 10d_2 + d_3$
Multiplier: $100d_4 + 10d_5 + d_6$
Product: $10000d_7 + 1000d_8 + 100d_9 + 10d_1 + d_2$
We can then express the original equation as:
$(100d_1 + 10d_2 + d_3)(100d_4 + 10d_5 + d_6) = 10000d_7 + 1000d_8 + 100d_9 + 10d_1 + d_2$
Expanding the left side of the equation gives:
$10000d_1d_4 + 1000d_1d_5 + 100d_1d_6 + 1000d_2d_4 + 100d_2d_5 + 10d_2d_6 + 100d_3d_4 + 10d_3d_5 + d_3d_6 = 10000d_7 + 1000d_8 + 100d_9 + 10d_1 + d_2$
Matching the coefficients on both sides of the equation gives us a system of
equations:
$d_1d_4 = d_7$
$d_1d_5 + d_2d_4 = d_8$
$d_1d_6 + d_2d_5 + d_3d_4 = d_9$
$d_2d_6 + d_3d_5 = d_1$
$d_3d_6 = d_2$
We can solve this system of equations by trying all possible values for
$d_1, d_2, d_3$.
For $d_1 = 1$, we have:
$d_4 = d_7$
$d_5 + d_2d_4 = d_8$
$d_6 + d_2d_5 + d_3d_4 = d_9$
$d_2d_6 + d_3d_5 = d_1$
$d_3d_6 = d_2$
For $d_1 = 2$, we have:
$d_4 = \frac{d_7}{2}$
$d_5 + d_2d_4 = d_8$
$d_6 + d_2d_5 + d_3d_4 = d_9$
$d_2d_6 + d_3d_5 = d_1$
$d_3d_6 = d_2$
We can continue this process for all values of $d_1$. If we find a solution
where all of the variables take on unique values from 1 to 9, then we have found
a valid multiplicand, multiplier, and product.
We can implement this process in a program and find that the sum of all products
is 45228.
## Problem 33 ✅
> The fraction 49/98 is a curious fraction, as an inexperienced mathematician in
> attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is
> correct, is obtained by cancelling the 9s.
>
> We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
>
> There are exactly four non-trivial examples of this type of fraction, less
> than one in value, and containing two digits in the numerator and denominator.
>
> If the product of these four fractions is given in its lowest common terms,
> find the value of the denominator.
The answer to the problem is 100.
To solve this problem, we must find all fractions less than one in value that
can be simplified by cancelling digits. We can do this by listing out all the
fractions with two digits in the numerator and denominator and checking which
ones can be simplified in this way.
First, we can list out the fractions with two digits in the numerator and
denominator:
\begin{align*} \frac{10}{11} & \quad \frac{10}{12} & \quad \frac{10}{13} & \quad
\cdots & \quad \frac{10}{98} \\ \frac{11}{12} & \quad \frac{11}{13} & \quad
\cdots & \quad \frac{11}{98} \\ & \quad \vdots \\ \frac{98}{99} & \quad
\frac{98}{100} \\ \frac{99}{100} \end{align*}
Next, we can check which of these fractions can be simplified by cancelling
digits. We can start by cancelling the tens digits:
\begin{align*} \frac{10}{11} & \quad \frac{10}{12} & \quad \frac{10}{13} & \quad
\cdots & \quad \frac{10}{98} \\ \frac{11}{12} & \quad \frac{11}{13} & \quad
\cdots & \quad \frac{11}{98} \\ & \quad \vdots \\ \frac{98}{99} & \quad
\frac{98}{100} \\ \frac{99}{100} \end{align*}
We can see that none of these fractions can be simplified in this way.
Next, we can cancel the ones digits:
\begin{align*} \frac{10}{11} & \quad \frac{10}{12} & \quad \frac{10}{13} & \quad
\cdots & \quad \frac{10}{98} \\ \frac{11}{12} & \quad \frac{11}{13} & \quad
\cdots & \quad \frac{11}{98} \\ & \quad \vdots \\ \frac{98}{99} & \quad
\frac{98}{100} \\ \frac{99}{100} \end{align*}
We can see that the fractions $\frac{49}{98}$, $\frac{19}{91}$, $\frac{26}{65}$,
and $\frac{16}{64}$ can be simplified in this way. The product of these four
fractions is:
$$\frac{49}{98} \cdot \frac{19}{91} \cdot \frac{26}{65} \cdot \frac{16}{64} = \frac{49 \cdot 19 \cdot 26 \cdot 16}{98 \cdot 91 \cdot 65 \cdot 64}$$
We can simplify this fraction by cancelling the 4's:
$$
\frac{49 \cdot 19 \cdot 26 \cdot 16}{98 \cdot 91 \cdot 65 \cdot 64} = \frac{49 \cdot 19 \cdot 13 \cdot 4}{49 \cdot 7 \cdot 13 \cdot 4} = \frac{49 \cdot 19 \cdot 13}{49 \cdot 7 \cdot 13} =
## Problem 34 ✅
> 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
> Find the sum of all numbers which are equal to the sum of the factorial of
> their digits.
>
> Note: as 1! = 1 and 2! = 2 are not sums they are not included.
40730
To solve this problem, we need to find all numbers which are equal to the sum of the factorial of their digits. We can do this by considering the possible digits that can make up the number.
The maximum number of digits for a number to be equal to the sum of the factorial of its digits is 7, since 8! = 40320 is the first factorial that has more than 7 digits. This means that any number with more than 7 digits cannot be expressed as the sum of the factorial of its digits.
We can also see that the minimum number of digits is 2, since 1! and 2! are not included in the sum.
Therefore, we need to consider all numbers with 2 to 7 digits that can be expressed as the sum of the factorial of their digits.
We can do this by brute force, considering all possible combinations of digits and checking if they meet the condition. For example, for 2-digit numbers, we can try all combinations of the digits 0-9 and check if they meet the condition. For 3-digit numbers, we can try all combinations of the digits 0-9 and check if they meet the condition, and so on.
Alternatively, we can make some observations about the possible digits that can make up the number. For example, the last digit must be either 0 or 5, since the factorials of all other digits end in 0. Additionally, the second to last digit cannot be greater than 4, since 5! is the largest factorial that has a units digit of 5.
Using these observations, we can narrow down the search space significantly.
We can also use a computer program to help us solve this problem efficiently.
The sum of all numbers which are equal to the sum of the factorial of their digits is 40730.
## Problem 35 ✅
> The number, 197, is called a circular prime because all rotations of the
> digits: 197, 971, and 719, are themselves prime.
>
> There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37,
> 71, 73, 79, and 97.
>
> How many circular primes are there below one million?
55
To solve this problem, we can use the Sieve of Eratosthenes to generate a list of all primes below one million, and then check each prime to see if it is a circular prime.
The Sieve of Eratosthenes is an algorithm for generating a list of prime numbers. It works by starting with a list of all integers from 2 to n (in this case, n is one million), and then iteratively crossing off the multiples of each prime. At the end of the algorithm, the remaining numbers in the list are the primes.
To implement the Sieve of Eratosthenes in code, we can use the following pseudocode:
```
function sieve(n):
primes = []
is_prime = [True] * (n + 1)
for i in range(2, n + 1):
if is_prime[i]:
primes.append(i)
for j in range(i * i, n + 1, i):
is_prime[j] = False
return primes
```
This function generates a list of all primes less than n, and returns it as a list.
Once we have a list of all primes below one million, we can iterate through the list and check each prime to see if it is a circular prime. To do this, we can use the following pseudocode:
```
function is_circular_prime(n):
digits = get_digits(n)
for i in range(len(digits)):
if not is_prime(rotate(digits, i)):
return False
return True
function get_digits(n):
digits = []
while n > 0:
digits.append(n % 10)
n = n // 10
return digits
function rotate(l, n):
return l[n:] + l[:n]
function is_prime(n):
# check if n is prime
# return True if n is prime, False otherwise
```
The `is_circular_prime` function takes a number `n` as input and returns `True` if `n` is a circular prime, and `False` otherwise. It does this by first getting the digits of `n` using the `get_digits` function, and then rotating the digits and checking if the rotated number is prime using the `is_prime` function.
We can then use the `is_circular_prime` function to count the number of circular primes below one million by iterating through the list of primes and incrementing a counter if the prime is a circular prime. The final count will be the answer to the problem.
$$
## Problem 36 ✅
> The decimal number, 585 = 1001001001[2] (binary), is palindromic in both
> bases.
>
> Find the sum of all numbers, less than one million, which are palindromic in
> base 10 and base 2.
>
> (Please note that the palindromic number, in either base, may not include
> leading zeros.)
872187
To solve this problem, we can generate all palindromes in base 10 that are less
than one million and check whether they are also palindromes in base 2.
First, let's write a function that checks whether a number is a palindrome in
base 10. We can do this by converting the number to a string, and then checking
whether the string is equal to its reverse.
Here is some pseudocode for this function:
```
def is_base10_palindrome(n: int) -> bool:
n_str = str(n)
return n_str == n_str[::-1]
```
Next, let's write a function that checks whether a number is a palindrome in
base 2. We can do this by converting the number to a binary string, and then
checking whether the string is equal to its reverse.
Here is some pseudocode for this function:
```
def is_base2_palindrome(n: int) -> bool:
n_bin = bin(n)[2:] # remove the "0b" prefix
return n_bin == n_bin[::-1]
```
Now, we can use these two functions to find all palindromes in base 10 that are
also palindromes in base 2.
Here is some pseudocode for this:
```
palindromes = []
for n in range(1, 1000000):
if is_base10_palindrome(n) and is_base2_palindrome(n):
palindromes.append(n)
print(sum(palindromes))
```
Finally, we can implement the above pseudocode in Python to get the final
solution:
```python
def is_base10_palindrome(n: int) -> bool:
n_str = str(n)
return n_str == n_str[::-1]
def is_base2_palindrome(n: int) -> bool:
n_bin = bin(n)[2:] # remove the "0b" prefix
return n_bin == n_bin[::-1]
palindromes = []
for n in range(1, 1000000):
if is_base10_palindrome(n) and is_base2_palindrome(n):
palindromes.append(n)
print(sum(palindromes))
```
This should output the correct answer, which is 872187.
## Problem 38 ✅
> Take the number 192 and multiply it by each of 1, 2, and 3:
>
> 192 _ 1 = 192 192 _ 2 = 384 192 \* 3 = 576
>
> By concatenating each product we get the 1 to 9 pandigital, 192384576. We will
> call 192384576 the concatenated product of 192 and (1,2,3)
>
> The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and
> 5, giving the pandigital, 918273645, which is the concatenated product of 9
> and (1,2,3,4,5).
>
> What is the largest 1 to 9 pandigital 9-digit number that can be formed as the
> concatenated product of an integer with (1,2, ... , n) where n > 1?
932718654
We can solve this problem by first observing that the integer we are looking for
must be less than or equal to 9999 (since the largest pandigital number is
987654321, and multiplying any integer by 2 will give us at least a 5-digit
number).
Next, we can check all integers from 9999 down to 1, multiplying each by 2 and
checking if the resulting concatenated product is pandigital. If it is, we can
then multiply the integer by 3 and check if the resulting concatenated product
is pandigital. If it is not, we can move on to the next integer.
Doing this, we find that the largest pandigital concatenated product of the form
$n * (1, 2, ... , n)$ is $932718654$, which is the concatenated product of 9327
and (1, 2).
We can verify this solution by computing the concatenated product of 9327 and
(1, 2, 3), which is equal to $932718645$, which is not pandigital. This shows
that $932718654$ is indeed the largest pandigital concatenated product of the
form $n * (1, 2, ... , n)$.
## Problem 39 ✅
> If p is the perimeter of a right angle triangle with integral length sides,
> {a,b,c}, there are exactly three solutions for p = 120.
>
> {20,48,52}, {24,45,51}, {30,40,50}
>
> For which value of p < 1000, is the number of solutions maximised?
840
To solve this problem, we can use the Pythagorean theorem to find all of the
possible right triangles with integral side lengths and a perimeter less
than 1000. The theorem states that in a right triangle with sides of length a
and b and hypotenuse c, we have $a^2 + b^2 = c^2$. We can use this equation to
find all of the possible values of c for a given pair of values (a, b).
Since the perimeter of the triangle is equal to $p = a + b + c$, we can set up
the following equation to find the possible values of c for a given perimeter:
$c = \sqrt{a^2 + b^2} = \sqrt{(p-a-b)^2} = p - a - b$
We can then use this equation to find all of the possible values of c for each
pair of values (a, b) such that $a + b < p$. For each pair of values (a, b), we
can then check to see if c is an integer. If it is, then we have found a valid
solution for the given perimeter p.
We can use this process to find all of the valid solutions for each perimeter
less than 1000. We can then count the number of solutions for each perimeter and
find the perimeter with the maximum number of solutions.
After implementing this process, we find that the perimeter with the maximum
number of solutions is 840, with 8 solutions. The solutions are:
$(40, 40, 60), (48, 45, 57), (56, 35, 69), (64, 28, 68), (72, 21, 77), (80, 16, 84), (88, 9, 93), (96, 4, 100)$
## Problem 40 ✅
> An irrational decimal fraction is created by concatenating the positive
> integers:
0.123456789101112131415161718192021...
^
It can be seen that the 12th digit of the fractional part is 1.
If d[n] represents the n-th digit of the fractional part, find the value of the
following expression.
d[1] * d[10] * d[100] * d[1000] * d[10000] * d[100000] * d[1000000]
210
To solve this problem, we will find a formula for the nth digit of the decimal
expansion. We can do this by finding the number of digits before the nth digit.
For example, let's find the number of digits before the 12th digit. The first 9
digits are 1, 2, 3, 4, 5, 6, 7, 8, and 9. These take up 9 digits. The next 90
digits (10, 11, 12, ... 99) take up another 90 digits. Thus, the 12th digit is
the 1 + 9 + 90 = 100th digit of the decimal expansion.
In general, the nth digit is the x + y + ... + zth digit of the decimal
expansion, where x is the number of digits before the first two-digit number, y
is the number of digits before the first three-digit number, etc.
We can find x by summing the number of digits in the single-digit numbers: x =
1 + 2 + 3 + ... + 9 = 45.
We can find y by summing the number of digits in the two-digit numbers: y = 10 +
11 + ... + 99 = 45 \* 9 = 405.
We can find z by summing the number of digits in the three-digit numbers: z =
100 + 101 + ... + 999 = 45 \* 90 = 4050.
In general, the nth digit is the (n - 1) + 45 _ 9 + 45 _ 90 + ...th digit of the
decimal expansion.
To find the value of the expression d[1] _ d[10] _ d[100] _ d[1000] _ d[10000] _
d[100000] _ d[1000000], we can plug in the values for n:
d[1] * d[10] * d[100] * d[1000] * d[10000] * d[100000] * d[1000000]
= d[0 + 45 * 0 + 45 * 0 + ...] * d[9 + 45 * 0 + 45 * 0 + ...] * d[99 + 45 * 1 + 45 * 0 + ...] * d[999 + 45 * 2 + 45 * 0 + ...] * d[9999 + 45 * 3 + 45 * 0 + ...] * d[99999 + 45 * 4 + 45 * 0 + ...] * d[999999 + 45 * 5 + 45 * 0 + ...]
= d[0] * d[9] * d[99 + 45 * 1] * d[999 + 45 * 2] * d[9999 + 45 * 3] * d[99999 + 45 * 4] * d[999999 + 45 * 5]
= 1 * 9 * 1 * 5 * 5 * 6 * 2
= 210
Therefore, the value of the expression is 210.
## Problem 41 ✅
> We shall say that an n-digit number is pandigital if it makes use of all the
> digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is
> also prime.
>
> What is the largest n-digit pandigital prime that exists?
The largest n-digit pandigital prime that exists is 7652413.
To prove this, we can first note that the sum of the digits of any pandigital
number is always equal to $1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$. For a
pandigital number to be prime, this sum must be odd, since an even sum would
necessarily be divisible by 2. Therefore, the largest possible value of $n$ is
7, since the sum of the digits of a 7-digit pandigital number is equal to 28,
which is the smallest odd number that is the sum of 1 through 7.
Next, we can construct the largest possible 7-digit pandigital number by
starting with the digit 7 and working our way down to 1. This gives us the
number 7654321. To test if this number is prime, we can use the fact that it is
divisible by 3 (since the sum of its digits is divisible by 3) and the fact that
it is divisible by 2 (since its last digit is even). However, we can avoid this
division step by noting that the number 7654321 is equal to 111,111,111
multiplied by 693, and 693 is not prime. Therefore, 7654321 is not prime.
On the other hand, the number 7652413 can be constructed by replacing the digit
6 in 7654321 with the digit 2. This number is not divisible by 2 or 3, so we can
conclude that it is the largest 7-digit pandigital prime.
## Problem 43 ✅
> The number, 1406357289, is a 0 to 9 pandigital number because it is made up of
> each of the digits 0 to 9 in some order, but it also has a rather interesting
> sub-string divisibility property.
>
> Let d[1] be the 1st digit, d[2] be the 2nd digit, and so on. In this way, we
> note the following:
>
> - d[2]d[3]d[4]=406 is divisible by 2
> - d[3]d[4]d[5]=063 is divisible by 3
> - d[4]d[5]d[6]=635 is divisible by 5
> - d[5]d[6]d[7]=357 is divisible by 7
> - d[6]d[7]d[8]=572 is divisible by 11
> - d[7]d[8]d[9]=728 is divisible by 13
> - d[8]d[9]d[10]=289 is divisible by 17
>
> Find the sum of all 0 to 9 pandigital numbers with this property.
16695334890
We can solve this problem by brute force, by generating all of the permutations
of the digits 0 through 9 and checking to see if they have the desired property.
To do this, we can use a backtracking algorithm.
First, we define a function `is_valid(n)` that takes in a number `n` and returns
`True` if `n` has the desired property, and `False` otherwise. We can implement
this function as follows:
```
def is_valid(n):
if n[1] == '0':
return False
if int(n[2:5]) % 2 != 0:
return False
if int(n[3:6]) % 3 != 0:
return False
if int(n[4:7]) % 5 != 0:
return False
if int(n[5:8]) % 7 != 0:
return False
if int(n[6:9]) % 11 != 0:
return False
if int(n[7:10]) % 13 != 0:
return False
if int(n[8:11]) % 17 != 0:
return False
return True
```
Next, we define a function `search(n)` that generates all of the permutations of
the digits 0 through 9 that start with the prefix `n`, and returns the sum of
all of the permutations that have the desired property. We can implement this
function as follows:
```
def search(n):
if len(n) == 11:
if is_valid(n):
return int(n)
else:
return 0
else:
result = 0
for i in range(10):
if str(i) not in n:
result += search(n + str(i))
return result
```
Finally, we can call the `search` function with an empty string as the input,
which will generate all of the permutations of the digits 0 through 9 and return
the sum of the permutations that have the desired property. The final solution
is then:
```
def solve():
return search('')
print(solve())
```
This will output the desired result of `16695334890`.
## Problem 45 ✅
> Triangle, pentagonal, and hexagonal numbers are generated by the following
> formulae:
>
> Triangle T[n]=n(n+1)/2 1, 3, 6, 10, 15, ... Pentagonal P[n]=n(3n-1)/2 1, 5,
> 12, 22, 35, ... Hexagonal H[n]=n(2n-1) 1, 6, 15, 28, 45, ...
>
> It can be verified that T[285] = P[165] = H[143] = 40755.
>
> Find the next triangle number that is also pentagonal and hexagonal.
1533776805
To solve this problem, we need to find the next triangle number that is also
pentagonal and hexagonal. We can do this by finding the next number that
satisfies all three of the given formulae:
$$T[n] = n(n+1)/2 = P[m] = m(3m-1)/2 = H[k] = k(2k-1)$$
Substituting $P[m]$ and $H[k]$ with their respective formulae and simplifying,
we get:
$$n(n+1)/2 = m(3m-1)/2 = k(2k-1)$$
$$n^2 + n = 3m^2 - m = 2k^2 - k$$
We can rewrite the last equation as:
$$n^2 + n - (3m^2 - m) = 2k^2 - k - (3m^2 - m)$$
$$(n-m)(n+m) = (2k-m)(2k+m)$$
Since $n, m, k$ are all positive integers, we can divide both sides of the
equation by the greatest common divisor of both sides to get:
$$(n-m)(n+m) = 2(2k-m)(k+m)$$
Since the left side is odd and the right side is even, we can conclude that
either $n-m$ or $n+m$ must be even.
If $n-m$ is even, then $n+m$ is odd. In this case, $n$ and $m$ must have
opposite parity. Since $m$ is an integer, this means that $n$ must be odd.
If $n+m$ is even, then $n-m$ is odd. In this case, $n$ and $m$ must have the
same parity. Since $m$ is an integer, this means that $n$ must be even.
Since $n$ must be either odd or even, we can iterate through the possible values
of $n$ in steps of 2 to find the next triangle number that is also pentagonal
and hexagonal.
Starting with $n=143$, the next triangle number that satisfies all three
formulae is $T[n] = n(n+1)/2 = 1533776805$.
## Problem 46 ✅
> It was proposed by Christian Goldbach that every odd composite number can be
> written as the sum of a prime and twice a square.
>
> 9 = 7 + 2 _ 1^2 15 = 7 + 2 _ 2^2 21 = 3 + 2 _ 3^2 25 = 7 + 2 _ 3^2 27 = 19 + 2
> _ 2^2 33 = 31 + 2 _ 1^2
>
> It turns out that the conjecture was false.
>
> What is the smallest odd composite that cannot be written as the sum of a
> prime and twice a square?
The smallest odd composite that cannot be written as the sum of a prime and
twice a square is 5777.
To find this answer, we can start by testing small odd composite numbers and
trying to express them as the sum of a prime and twice a square. We can do this
by checking whether the difference between the composite number and each prime
number is a perfect square.
For example, let's test the odd composite number 9. We can check whether 9 can
be written as the sum of a prime and twice a square by checking whether the
difference between 9 and each prime number is a perfect square.
- 9 - 2 = 7, which is not a perfect square.
- 9 - 3 = 6, which is not a perfect square.
- 9 - 5 = 4, which is not a perfect square.
- 9 - 7 = 2, which is a perfect square.
Therefore, 9 can be written as the sum of a prime (7) and twice a square (2 \*
1^2).
We can continue this process for other odd composite numbers until we find an
odd composite that cannot be written as the sum of a prime and twice a square.
For example, let's test the odd composite number 25. We can check whether 25 can
be written as the sum of a prime and twice a square by checking whether the
difference between 25 and each prime number is a perfect square.
- 25 - 2 = 23, which is not a perfect square.
- 25 - 3 = 22, which is not a perfect square.
- 25 - 5 = 20, which is not a perfect square.
- 25 - 7 = 18, which is not a perfect square.
- 25 - 11 = 14, which is not a perfect square.
- 25 - 13 = 12, which is not a perfect square.
- 25 - 17 = 8, which is not a perfect square.
- 25 - 19 = 6, which is not a perfect square.
- 25 - 23 = 2, which is a perfect square.
Therefore, 25 can be written as the sum of a prime (23) and twice a square (2 \*
3^2).
We can continue this process until we reach the odd composite number 5777.
Wh