Project Eulers 7 and 8
Just a few more Euler’s left and I realized the other day how long it had been since I’d posted some answers! I’m also posting both 7 and 8 because really, problem 7 uses the prime library again, so the answer is really short and sweet.
So, first, here is the problem: 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 10 001st prime number?
First, here are the tests:
require 'problem7/problem7'
describe 'Prime number positions' do
it "is the 6th prime number" do
expect(Problem7.prime_place(6)).to eq 13
end
it "is the 10001st prime number" do
expect(Problem7.prime_place(10001)).to eq 104743
end
end
The tests and the code are pretty simple. You’re just using the resources in the prime library and then use the library to list (take) the numbers up until a certain position (ie- the 10,001st position) and then put the last number which is the answer to the question. So, here’s the code:
require 'prime'
module Problem7
def self.prime_place(position)
prime_place = Prime.take(position).last
end
puts Prime.take(10001).last
end
And now for Euler 8. This problem was actually really tough for me for two reasons… first, it seemed different than most of the others I had done up until now and second, how the heck do you test this thing?!
Here’s the problem:
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
So, first for the test. After asking around a bit, the best suggestion I got for testing was to break down the string and take 10 or 15 characters and figure out the largest product from that string and then do the same with the larger number.
Here are the tests:
require 'problem8/problem8'
describe 'largest products of consecutive numbers' do
it "is the largest product of 5 consective numbers" do
expect(Problem8.product(7316717653)).to eq 1764
end
it "is the largest product of 5 consecutive numbers" do
expect(Problem8.product(7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450)).to eq 40824
end
end
And so, here’s the solution. First, I created an empty array. Then I wanted to use
.each_cons
which takes every set of consecutive numbers based on the number of characters you ask for (in this case, it would be 5 because I’m looking for the largest product of 5 consecutive numbers) but .each_cons
wouldn’t work because you can’t call .each_cons
on a string. So, first, I had to separate the string into individual characters by using each_char
. Once the string was separated into each_char
(each character) I used the map method to make each of the string characters into an array of integers. Then, I used each_cons(5)
which separated the array of integers into arrays of every five characters. The I took the product of each of those integers and pushed it into an array. Finally, the max is called on that array which gives the largest number needed for the answer.module Problem8
def self.product
arr = []
"731671765313306249192251196744265747423553491949349698352"<br /> "0312774506326239578318016984801869478851843858615607891129"<br /> "4949545950173795833195285320880551112540698747158523863050"<br /> "7156932909632952274430435576689664895044524452316173185640"<br /> "3098711121722383113622298934233803081353362766142828064444"<br /> "8664523874930358907296290491560440772390713810515859307960"<br /> "8667017242712188399879790879227492190169972088809377665727"<br /> "3330010533678812202354218097512545405947522435258490771167"<br /> "0556013604839586446706324415722155397536978179778461740649"<br /> "5514929086256932197846862248283972241375657056057490261407"<br /> "9729686524145351004748216637048440319989000889524345065854"<br /> "1227588666881164271714799244429282308634656748139191231628"<br /> "2458617866458359124566529476545682848912883142607690042242"<br /> "1902267105562632111110937054421750694165896040807198403850"<br /> "9624554443629812309878799272442849091888458015616609791913"<br /> "3875499200524063689912560717606058861164671094050775410022"<br /> "5698315520005593572972571636269561882670428252483600823257"<br /> "530420752963450".each_char.map(&:to_i).each_cons(5) { |a| p arr << a.reduce(:*) }
puts arr.max
end
end
Phew! Look forward to the last two Eulers, 9 and 10, which I’ll hopefully get to post soon.