# Project Euler: Problem 8

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

## Solution

This one wasn’t too complicated to do, and there isn’t really that much optimizations or super fancy math stuff that can help either.

Simply put the numbers in an array, and walk through it.

var d = ("73167176531330624919225119674426574742355349194934"
+ "96983520312774506326239578318016984801869478851843"
+ "85861560789112949495459501737958331952853208805511"
+ "12540698747158523863050715693290963295227443043557"
+ "66896648950445244523161731856403098711121722383113"
+ "62229893423380308135336276614282806444486645238749"
+ "30358907296290491560440772390713810515859307960866"
+ "70172427121883998797908792274921901699720888093776"
+ "65727333001053367881220235421809751254540594752243"
+ "52584907711670556013604839586446706324415722155397"
+ "53697817977846174064955149290862569321978468622482"
+ "83972241375657056057490261407972968652414535100474"
+ "82166370484403199890008895243450658541227588666881"
+ "16427171479924442928230863465674813919123162824586"
+ "17866458359124566529476545682848912883142607690042"
+ "24219022671055626321111109370544217506941658960408"
+ "07198403850962455444362981230987879927244284909188"
+ "84580156166097919133875499200524063689912560717606"
+ "05886116467109405077541002256983155200055935729725"
+ "71636269561882670428252483600823257530420752963450")
.Select(x => Convert.ToUInt64(x.ToString()))
.ToArray();

var maxProduct = 0UL;

for (int i = 0; i < d.Length - 4; i++)
{
var product = d[i]
* d[i + 1]
* d[i + 2]
* d[i + 3]
* d[i + 4];

if (product > maxProduct)
maxProduct = product;
}

Nothing fancy, but works nicely. Finishes in around 3 milliseconds.

How did you do it? Do you have a more clever or efficient way to do this? Please share ðŸ™‚

• Amir

perl version:

Everything before to

var product = d[i]
* d[i + 1]
* d[i + 2]
* d[i + 3]
* d[i + 4];

is fine, but if you run a check:

next if (d[i + 1] == 1 || d[i + 1] == 0;
next if (d[i + 2] == 1 || d[i + 2] == 0;
next if (d[i + 3] == 1 || d[i + 3] == 0;
next if (d[i + 4] == 1 || d[i + 4] == 0;

before doing the actual computation you can save a lot of time, since having 0 in a computation is pointless, and having a 1 will do you no good.

just a thought ðŸ™‚

• Interesting… but will that checking be cheaper than just doing the multiplication?

• Amir

In terms of speed i don’t know, you have to check, but i think it is.
In terms of logic its probably more sensible, why multiple numbers that are useless?

• Amir

I also think if you add the numbers instead of multiplying them you might save a couple of nano seconds, lol, and save memory(which in this case if probably close to nothing).