1996 AIME Problem 3

Attempt Problem 3 of the 1996 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1996 AIME solutions, or check the answer key.

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3.

Find the smallest positive integer nn for which the expansion of (xy3x+7y21)n,(xy-3x+7y-21)^n, after like terms have been collected, has at least 19961996 terms.

Answer: 44
Concepts:factoringbinomial theoremcounting pairs
Difficulty rating: 1690
Small Hint:

Factor the expression before raising it to the nnth power

Big Hint:

Count the distinct choices of the exponents of xx and yy

Solution:

The base factors as xy3x+7y21=(x+7)(y3).\begin{gathered}xy-3x+7y-21\\=(x+7)(y-3).\end{gathered} Hence its nnth power is (x+7)n(y3)n.(x+7)^n(y-3)^n. Each exponent of xx from 00 through nn can occur with each exponent of yy from 00 through n,n, and every resulting coefficient is nonzero. Thus there are (n+1)2(n+1)^2 terms. Since 442<1996452,44^2<1996\leq45^2, the least possible n+1n+1 is 45,45, so n=44.n=44.

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