One million is good round number :) Your task for the weekend challenge is to write code which check if you can find two numbers X and Y which product makes exactly 1 000 000, the catch is that none of these numbers should contain 0 inside it.

How fast is your code?

## The rules:

You can code the solution in any programming language during the weekend and have to submit it to info@olimex.com latest on Sunday July 27th.

On Monday we will upload the solutions on GitHub https://github.com/OLIMEX/WPC

You can play with your real name or with nick if you want to be anonymous, we will not disclosure your personal info if you do not want to.

max

Jul 25, 2014@ 18:33:041,000,000 = 10^6 = 5^6 * 2^6

All possible pair of positive integer factors are (5^m * 2^n, 5^(6-m) * 2^(6-n) ) with n,m = 0..6

BUT if you put a five with a two in a factor you get a multiple of 10, then you get a zero. Fives and twos must remain separate.

So the only remaining combination should be (5^6, 2^6) = (15625, 64)

Am I missing anything?

If you know that X * Y = 1,000,000 you only need to check if neither one is divisible by 10.

result = ! ( (X % 10 == 0) || (Y % 10 == 0) )

or, since there’s only 1 solution

result = (X == 64) || (Y == 64) [once we know that X*Y=1,000,000]

Edoardo Codeglia

Jul 26, 2014@ 00:40:11Also it’s worth to remember that there are actually four solutions :)

15625*64

64*15625

-64*-15625

-15625*-64

-(-) = + :)

max

Jul 26, 2014@ 11:54:23That’s why I wrote “positive integer factors” and “combination” (instead of permutation). Under those restrictions, the solution should be unique.

marameo00

Jul 25, 2014@ 23:47:59So now anybody submit his code. But this is sure the best solution. I thought there’s a solution like yours but I’m not able to find.

Is it possible to generalize the solution for generic X*Y=Z (all integers value) so that X nor Y cantains digit 0 (or even better digit d)?

marameo00

Jul 25, 2014@ 23:54:31…

marek

Jul 26, 2014@ 05:38:44Note 2^10 is 1024 and has a 0, so not being divisible by 10 doesn’t guarantee there isn’t a zero in the solution. That is if you’re writing a generalized solution which factors any number.

marek

Jul 26, 2014@ 05:47:397000000 makes an interesting test case, where all the 2 factors are grouped (64), all the 5 factors are grouped (15625) and the 7 factor needs to be assigned to (64).

My brute-force approach would be to factor Z, and create all possible combinations (non-duplicate) of the factors and see which ones have and don’t have 0’s.

Dylan

Jul 26, 2014@ 15:45:24It wont work for 3e10 = 3 * 2^10 * 5^10 (3072 * whatever).

Michael van Niekerk

Jul 28, 2014@ 06:32:06I don’t know how to submit – but https://github.com/mvniekerk/OlimexWeekend94 here’s my code.

Followed the brute force method mostly.