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):
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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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