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CIE 9231 2026 June Paper 12 Q2

A Level / CIE / FP1

CIE 9231 2026 June Paper 12 Paper · Question 2

题目

Problem

Prove by mathematical induction that 36n+92n+893^{6n}+92^n+89 is divisible by 9191 for all positive integers nn.

(6)

题目中文翻译

用数学归纳法证明:对于所有正整数 nn,36n+92n+893^{6n}+92^n+89 都能被 9191 整除。

(6)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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按归纳法完整写出基础步骤、归纳假设和由k到k+1的递推。利用3的6次方等于728+1,且728是91的倍数。

答题过程

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Let F(n)=36n+92n+89F(n)=3^{6n}+92^n+89.

For n=1n=1,

F(1)=729+92+89=910=91(10),F(1)=729+92+89=910=91(10),

so the statement is true.

Assume that F(k)F(k) is divisible by 9191 for some positive integer kk. Since 36=729=728+13^6=729=728+1 and 92=91+192=91+1,

F(k+1)= (728+1)36k+(91+1)92k+89= (36k+92k+89) +728⋅36k+91⋅92k= F(k)+91(8⋅36k+92k).\begin{align*} F(k+1)=&\,(728+1)3^{6k}+(91+1)92^k+89 \\[4mm] =&\,\big(3^{6k}+92^k+89\big)\\[4mm] &\,+728\cdot3^{6k}+91\cdot92^k \\[4mm] =&\,F(k)+91\big(8\cdot3^{6k}+92^k\big). \end{align*}

The first term is divisible by 9191 by the inductive hypothesis, and the second is explicitly a multiple of 9191. Hence F(k+1)F(k+1) is divisible by 9191.

Therefore, by mathematical induction, 36n+92n+89 is divisible by 91\boxed{3^{6n}+92^n+89\text{ is divisible by }91} for every positive integer nn.