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"\ "0312774506326239578318016984801869478851843858615607891129"\ "4949545950173795833195285320880551112540698747158523863050"\ "7156932909632952274430435576689664895044524452316173185640"\ "3098711121722383113622298934233803081353362766142828064444"\ "8664523874930358907296290491560440772390713810515859307960"\ "8667017242712188399879790879227492190169972088809377665727"\ "3330010533678812202354218097512545405947522435258490771167"\ "0556013604839586446706324415722155397536978179778461740649"\ "5514929086256932197846862248283972241375657056057490261407"\ "9729686524145351004748216637048440319989000889524345065854"\ "1227588666881164271714799244429282308634656748139191231628"\ "2458617866458359124566529476545682848912883142607690042242"\ "1902267105562632111110937054421750694165896040807198403850"\ "9624554443629812309878799272442849091888458015616609791913"\ "3875499200524063689912560717606058861164671094050775410022"\ "5698315520005593572972571636269561882670428252483600823257"\ "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.

## No comments:

## Post a Comment