Question

Given that n is a positive integer, and the equation 2x+2y+z=n has a total of 28 positive integer solutions for (x, y, z). Find the sum of all possible values of n.​

Answer 1

Explanation:

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Answer 2

Assume x,y,z are required to be positive integers.

Consider two cases . . .

Case (1):n is even.

From the equation 2x+2y+z=n, it follows that z is also even.

Writing z=2w, and n=2m, for some positive integers w,m, we get

⟺⟺2x+2y+z=n2x+2y+2w=2mx+y+w=m

By the Stars-and-Bars formula, the equation

x+y+w=m

has exactly (m−12) positive integer solution triples (x,,w), for any given positive integer value of m, hence

⟹⟹⟹⟹⟹⟹(m−12)=28(m−1)(m−2)2=28m2−3m+2=56m2−3m−54=0(m−9)(m+6)=0m=9n=18

Case (2):n is odd.

From the equation 2x+2y+z=n, it follows that z is also odd.

Writing z=2w−1, and n=2m−1, for some positive integers w,m, we get

⟺⟺2x+2y+z=n2x+2y+(2w−1)=2m−1x+y+w=m

As for case (1), since m is a positive integer such that the equation

x+y+w=m

has exactly 28 positive integer solution triples (x,y,w), it follows that m=9, hence n=2m−1=17.

Note:If x,y,z are only required to be nonnegative integers, then using arguments analogous to the ones given above, we get n=12 or n=11.