1996 AIME 第 3 题

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3.

求最小正整数 nn,使得 (xy3x+7y21)n(xy-3x+7y-21)^n 展开并合并同类项后至少有 19961996 项。

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.

答案:44
知识点:因式分解二项式定理数对计数
难度评级:1690
小提示:

先将底式因式分解,再取其 nn 次方

Factor the expression before raising it to the nnth power

大提示:

计算 xxyy 的指数有多少种不同搭配

Count the distinct choices of the exponents of xx and yy

解答:

底式可分解为 xy3x+7y21=(x+7)(y3)\begin{gathered}xy-3x+7y-21\\=(x+7)(y-3)\end{gathered}\text{。}因此它的 nn 次方为 (x+7)n(y3)n(x+7)^n(y-3)^nxx 的指数可取从 00nn 的每个值,而每个值都可与 yy00nn 的每个指数搭配,并且所得各项的系数都不为零。因此共有 (n+1)2(n+1)^2 项。由于 442<199645244^2<1996\leq45^2n+1n+1 的最小可能值为 4545,所以 n=44n=44

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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