A heap of a positive integer number of matches (that is, no broken matches) are divided into five groups.

If we take as many matches from the first group as there are in the second group and add them to the second, and then take as many from the second group as there are in the third group and add them to the third, and, so on ...... until finally, we take as many from the fifth group as there are in the first group and add them to the first group - the number of matches in each of the groups would be equal to the same positive integer.

What is the minimum number of matches in each group at the beginning?

The following is a solution (I have not proven it minimal but I think it is):

Let X(i) be the number in the ith pile.
If there are n piles X(1), X(2), ... X(n)

X(1) = 2^n + 2^(n-1) - 1
X(2) = 2^n - 1
X(3) = 2^n - 2
...
X(i) = 2^n - 2^(i-2)
X(i+1) = 2^n - 2^(i-1)
...
X(n) = 2^n - 2^(n-2)

Proof:
After step 1
X(1) = 2^n + 2^(n-1) - 1 - (2^n - 1)= 2^(n-1)
X(2) = 2(2^n - 1) = 2*2^n - 2

After step 2
X(2) = 2*2^n - 2 - (2^n - 2) = 2^n
X(3) = 2(2^n - 2) = 2*2^n - 4

After step i-1
X(i) = 2(2^n - 2^(i-2)) = 2*2^n - 2^(i-1)

After step i
X(i) = 2*2^n - 2^(i-1) - (2^n - 2^(i-1)) = 2^n

After step n-1
X(n) = 2(2^n - 2^(n-2)) = 2*2^n - 2^(n-1)

After step n
X(n) =2*2^n - 2^(n-1) - (2^(n-1)) = 2(2^n - 2^(n-1)) = 2(2^(n-1) = 2^n
X(1) = 2*2^(n-1) = 2^n

So every pile ends up with 2^n matches.

Posted by Jer on 2012-02-27 12:55:44