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Unknown Power Ponder (Posted on 2014-08-11) Difficulty: 3 of 5
Each of x, y and z are positive integers such that x+y+z=2006 and, x!*y!*z! = p*10q, where each of p and q is a positive integer and p is not divisible by 10.

What is the smallest value of q?

No Solution Yet Submitted by K Sengupta    
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Some Thoughts re(2): Paper and pencil solution (spoiler?) | Comment 3 of 4 |
(In reply to re: Paper and pencil solution (spoiler?) by Ady TZIDON)

I get 492 as well.

Basically the idea is that the trailing zeros of n! number one for each multiple of 5, plus an extra one for each multiple of a power of 5.

Since 625 can be analysed as 1*625, and 5*125, and 25*25, and 125*5, 625! has 156 trailing zeros. So 624! has 152.

Then 758 can be analysed as 1*625, and 6*125, and 30*25, and 151*5, so 758! has 188 trailing zeros.

When these factorials are multiplied together, the result will thus have 492 zeros.



  Posted by broll on 2014-08-12 08:17:35
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