## Project Euler Problem 12 Solution

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?

My solution in Ruby:

```def calc_divisors(num)
res=[1]
2.upto(Math.sqrt(num).floor) do |i|
if num % i == 0
res << i
end
end
res.reverse.each do |i|
res << num / i
end
res.uniq
end

triangle = 0
count = 0
while calc_divisors(triangle).count < 500
count += 1
triangle += count
end
puts triangle```

This code takes a little while to run, and so far, this is my favourite Project Euler Solution. I did originally try brute-forcing the divisors for each number, but as you can imagine, testing half of every possible candidate for numbers in the millions meant that the loops were getting too long. Interestingly, I never let that original method find the answer so instead I hit google, looking for a faster way. I discovered a lot about sieves and old math geniuses, but the the best example I found was using prime factorials. The function ‘calc_divisors’ in the above code seems to be the fastest way I could find to determine the factors of a specified Integer. If you think there is a faster way, I’d be interested to hear.

## Project Euler Problem 11 Solution

In the 2020 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 2020 grid?

My solution in Ruby:

```grid = [
[8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8],
[49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0],
[81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65],
[52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 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, 3, 45, 2, 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, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21],
[24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72],
[21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95],
[78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92],
[16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57],
[86, 56, 0, 48, 35, 71, 89, 7, 5, 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, 4, 89, 55, 40],
[4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66],
[88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69],
[4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36],
[20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16],
[20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54],
[1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48]
]

biggestproduct = 0
factors = []
direction = nil
#Horizontal
for i in 0..19
for j in 0..(19-3)
product = grid[i][j] * grid[i][j+1] * grid[i][j+2] * grid[i][j+3]
factors = [grid[i][j], grid[i][j+1], grid[i][j+2], grid[i][j+3]] if product > biggestproduct
direction = "Horizontal" if product > biggestproduct
biggestproduct = product if  product > biggestproduct
end
end
#Vertical
for i in 0..(19-3)
for j in 0..19
product = grid[i][j] * grid[i+1][j] * grid[i+2][j] * grid[i+3][j]
factors = [grid[i][j], grid[i+1][j], grid[i+2][j], grid[i+3][j]] if product > biggestproduct
direction = "Vertical" if product > biggestproduct
biggestproduct = product if  product > biggestproduct
end
end
#Diagonal \
for i in 0..(19-3)
for j in 0..(19-3)
product = grid[i][j] * grid[i+1][j+1] * grid[i+2][j+2] * grid[i+3][j+3]
factors = [grid[i][j], grid[i+1][j+1], grid[i+2][j+2], grid[i+3][j+3]] if product > biggestproduct
direction = "Diagonal \\" if product > biggestproduct
biggestproduct = product if  product > biggestproduct
end
end
#Diagonal /
for i in 0..(19-3)
for j in (0+3)..19
product = grid[i][j] * grid[i+1][j-1] * grid[i+2][j-2] * grid[i+3][j-3]
factors = [grid[i][j], grid[i+1][j-1], grid[i+2][j-2], grid[i+3][j-3]] if product > biggestproduct
direction = "Diagonal /" if product > biggestproduct
biggestproduct = product if  product > biggestproduct
end
end
puts "answer: #{factors[0]} x #{factors[1]} x #{factors[2]} x #{factors[3]} = #{biggestproduct} (#{direction})"```

## Project Euler Problem 10 Solution

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

My solution in Ruby:

```def is_prime ( p )
if p == 2
return true
elsif p <= 1 || p % 2 == 0
return false
else
(3 .. Math.sqrt(p)).step(2) do |i|
if p % i == 0
return false
end
end
return true
end
end

sum = 0
for i in 2..2000000
if is_prime(i)
sum += i
end
end
puts sum```

## Project Euler Problem 9 Solution

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a2 + b2 = c2

paroxetine 40mgs price

For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

My solution in Ruby:

```def get_hypotenuse(a, b)
Math.sqrt(a**2 + b**2)
end

a = 1
b = 1
while (a < 1000) do
while ((a**2 + b**2) < 1000**2) do
c = get_hypotenuse(a, b)
if (a + b + c) == 1000
puts a*b*c
exit(0)
end
b += 1
end
b = 1
a += 1
end```

## Project Euler Problem 8 Solution

Find the greatest product of five 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

My solution in Ruby:

```var = %&
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450&

arr = var.split( // )
arr.delete "\n"
big = 0;
for i in 0..arr.length - 5
next if Integer(arr[i]) == 0
tmp = Integer(arr[i])
1.upto(4) { |j| tmp = tmp * Integer(arr[i + j]) }
big = tmp if tmp > big
end
puts big```

## Project Euler Problem 7 Solution

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6^(th) prime is 13.

What is the 10001^(st) prime number?

My solution in Ruby:

```def is_prime ( p )
if p == 2
return true
elsif p <= 1 || p % 2 == 0
return false
else
(3 .. Math.sqrt(p)).step(2) do |i|
if p % i == 0
return false
end
end
return true
end
end

prime_count = 6
prime_number = 13
number = 13
while prime_count < 10001 do
number += 2
if is_prime(number)
prime_count += 1
prime_number = number
end
end
puts '***********'
puts "#{prime_count}: #{prime_number}"```

## Project Euler Problem 6 Solution

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.

My solution in Ruby:

```def sum_and_square(j, k)
tmp = 0
for i in 1..100
tmp += i**j
end
tmp**k
end

puts sum_and_square(1,2) - sum_and_square(2,1)```

## Project Euler Problem 5 Solution

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?

My solution in Ruby:

```def has_remainder?(var)
1.upto(20) { |i| return true if var % i != 0 }
false
end

number = 0
loop do
number += 1
break if !has_remainder?(number)
end
puts number```

## Project Euler Problem 4 Solution

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.

My solution in Ruby:

```var = 0
999.downto(100) { |i|
999.downto(100) { |j|
prod = i * j
var = [var, prod].max if prod.to_s == prod.to_s.reverse
}
}
puts "Maximum palindrome is #{var}."```

## Project Euler Problem 3 Solution

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

My solution in Ruby:

```def is_prime ( p )
if p == 2
return true
elsif p <= 1 || p % 2 == 0
return false
else
(3 .. Math.sqrt(p)).step(2) do |i|
if p % i == 0
return false
end
end
return true
end
end

i, num = 1, 1
loop do
i += 1
if 600851475143 % i == 0
num = 600851475143 / i
break if is_prime(num)
end
end

puts "largest prime factor: " + num.to_s```