SOLUTION: Let n be a positive integer. If the equation 2x + 2y + z = n has 28 solutions in positive integers x, y and z, then what is the value of n?

Question 1148532: Let n be a positive integer. If the equation 2x + 2y + z = n has 28 solutions in positive integers x, y and z, then what is the value of n?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Find the answers (there are two of them) by looking at the numbers of solutions for particular values of n.

Suppose, for example, that n is 25. Then we have

2(x+y) is even, and 25 is odd, so z must be odd.

If z=1, then 2(x+y)=25-1=24, x+y=12, and there are 11 solutions in positive integers (x can be any integer from 1 to 11 inclusive).

If z=3, then 2(x+y)=25-3=22, x+y=11, and there are 10 solutions in positive integers.

If z=5, then 2(x+y)=25-5=20, x+y=10, and there are 9 solutions in positive integers.

The pattern should be clear. For n=25, the number of solutions in positive integers is 11+10+9+...+2+1 = 66.

We want the number of solutions to be 28, which is 1+2+3+4+5+6+7.

According to the pattern seen above, that will happen when z=1 and x+y=8; and that will be when n = 2(x+y)+z = 17.

A similar investigation with n being even shows that the number of solutions will also be 28 when n is 18.

ANSWERS: n is either 17 or 18 when the equation has 28 solutions.

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