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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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Solution Paper and pencil solution (spoiler?) | Comment 1 of 4
The problem might as well say  x!*y!*z! = p*5q , because there are more than enough 2's for each factor of 5.  The objective, I think is to minimize the total factors of 625, and then the total factors of 125 that are not a multiple of 625, and then the total factors of 25 that are not a multiple of 125, and then the total factors of 5 that are not a multiple of 25.

If two of the numbers are right below 625, I think that is one way of achieving the minimum.  I am thinking about 624 + 624 + 758.

If I have counted right,
5! is a multiple of 5^1 
25! is a multiple of 5^6  (because 5*1+ 1 = 6)
125! is a multiple of 5^31 (because 5*6+ 1 = 31)
625! is a multiple of 5^156 (because 5*31 + 1 = 156)
so 624! is a multiple of 5^152

758! is a multiple of 5^184 (ie, 156 + 31 - 3)

So the minimum q, I think is 152 + 152 + 184 = 488

Of course, I may have made a mistake ...

/**********************************/

Yes, I did make a mistake!

758! is a multiple of 5^188 (ie, 156 + 31 +1),
because 758 = 625 + 125 + 8

So the minimum q, is 152 + 152 + 188492

Edited on August 12, 2014, 10:52 am
  Posted by Steve Herman on 2014-08-11 21:01:22

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