An integer N consisting of five distinct nonzero digits has the curious property that it is equal to the sum of all the different three-digit integers formed by the three digit permutations of its five digits. Find N.

I missed this problem when it came out. Here is my take on the problem:

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Charlie's beginning of the explanation is how I would approach it:

"The way the solution can be achieved is to realize that in the sum of the permutations of three digits out of the five, each of the digits appears 12 times in each of the three positions (hundreds, tens, units).  The reason for that is that there are 4 choices for the first of the other two positions and 3 choices for the second.  As a result of its appearing 12 times in each of these positions, the total is 12*111*(a+b+c+d+e) where the letters are the digits making up the 5-digit number"  

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a + b + c + d + e could be (1+2+3+4+5) or 15 at least, and (5+6+7+8+9) or 35 at most. So that narrows things down somewhat, but, 12*111*(a+b+c+d+e) equals "abcde" but this can also be written as 9 * 148 * (a + b + c + d + e), and this means "abcde" is a multiple of 9, meaning the digits have to add up to nine. This means you need to check 18 and 27 as (a + b + c + d + e) and 18 * 1332 = 23976 which doesn't work, but 27 * 1332 = 35964

Posted by Gamer on 2004-07-01 12:36:20