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Project Euler Problem 5, 8 #2172

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LoopGlitch26 opened this issue Apr 9, 2021 · 2 comments
Open

Project Euler Problem 5, 8 #2172

LoopGlitch26 opened this issue Apr 9, 2021 · 2 comments
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@LoopGlitch26
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@LoopGlitch26 LoopGlitch26 commented Apr 9, 2021

Problem 8
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

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

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

Proposed Solution
Consider the number as a String and store it in a variable (Say S). And then loop through S.length() - 13 times. And then another loop will iterate through inside, to check the greatest product by using Integer.parseInt() and S.substring().

Answer
Answer:23514624000


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 positive number that is evenly divisible by all of the numbers from 1 to 20?

Proposed Solution
Starting from 11 till 20, run the loop to find the LCM in every iteration w.r.t the loop counter, and LCM is found using GCD function.

Answer
Answer:232792560

Alternate Solution
Initialize a variable to store integer value 20000, iterating through 20 skips, while checking for divisibility for all the numbers from 11 to 20

@LoopGlitch26
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@LoopGlitch26 LoopGlitch26 commented Apr 9, 2021

Please find the corresponding PR HERE

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@stale stale bot commented Jun 3, 2021

This issue has been automatically marked as stale because it has not had recent activity. It will be closed if no further activity occurs. Thank you for your contributions.

@stale stale bot added the stale label Jun 3, 2021
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