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Can someone please solve following problem.

Show that $9^n-2^n$ for any $n$ natural number is divisible by $7$. ($9 ^ n$ = $9$ to the power of $n$).

I know the principle of induction but am stuck with setting up a formula for this.

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5 Answers 5

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Hint:

Assume that $7\mid (9^n-2^n)$,

For $n+1:$

$$9^{n+1}-2^{n+1}=9^n9-2^n2=9^n(7+2)-2^n2=9^n7+9^n2-2^n2=9^n7+2(9^n-2^n)$$

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Let $f(k)=9^k-2^k$ where $k \in \mathbb{N}$.

Base Case: $f(1)=9-2=7=7(1)$.

Inductive Hypothesis: Assume $f(k)$ is divisible by $7$ for some natural number $k$.

$f(k+1)= 9^{k+1}-2^{k+1} = 9\cdot9^k-2\cdot2^k$.

Inductive step: $f(k+1)-2f(k)=7\cdot9^k=7\cdot9^k$. So $f(k+1) =7\cdot 9^k+2f(k)$.

Recall: $f(k)$ was assumed to be divisible by $7$ for the inductive hypothesis. So we are done since $f(k+1)$ is the sum of two things divisible by $7$.

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  • $\begingroup$ It would be clearer as follows: $f(k+1)-f(k)=\cdots=7\cdot9^k+f(k)$. $\endgroup$
    – lhf
    Commented Jun 1, 2021 at 14:40
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Hint: $x^n-y^n=(x-y)(x^{n-1}+yx^{n-2}+\dotsb+y^{n-2}x+y^{n-1})$ for $n\geq1$.

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    $\begingroup$ I'm not sure this is relevant to induction. $\endgroup$
    – pjs36
    Commented Mar 5, 2016 at 18:27
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Assume $7\mid(9^n-2^n)$; then you can write $9^n-2^n=7m$, for some integer $m$, or $9^n=7m+2^n$ as well. Then $$ 9^{n+1}-2^{n+1}= 9\cdot 9^n-2^{n+1}= 9(7m+2^n)-2^{n+1}= 63m+(9-2)2^n= 7(9m+2^n) $$

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There shouldn't really be a formula for this problem, just setting up the induction step which usually involves plugging in $n = k+1$ into your expression. First show $7\mid 9^n-2^n$ for $n=1$. Then suppose the result holds for some $k \geq 1$ (i.e. you are allowed to assume $9^k-2^k$ is divisible by $7$.) Consider then the quantity $9^{k+1}-2^{k+1}$. Proving that this quantity must also be divisible by $7$ is all that remains. Doing a little bit of algebra manipulation will get the result. For example notice that $$9^{k+1}-2^{k+1} = 9^{k}(7+2)-2^{k+1}$$ Can you proceed from here?

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