# Divisible by 7

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• Aug 21st 2010, 04:09 PM
Plato
I might as well add to this.
Suppose that it is true for $K$, i.e. $3^{2K+1}+2^{K+2}}$ is divisible by $7$.
Look at $3^{2(K+1)+1}+2^{(K+1)+2}=3^{2K+3}+2^{K+3}$
So $3^{2K+3}+2^{K+3}=3^{2K+3}+3^2
\cdot 2^{K+2}-3^2\cdot 2^{K+2}+2^{K+2}$
$=3^{2}\left(3^{2K+1}+2^{K+3}\right)-2^{K+2}\left(3^2-2\right)$.
That is clearly divisible by $7$.
• Aug 21st 2010, 04:21 PM
melese
Quote:

Originally Posted by Plato
I might as well add to this.
Suppose that it is true for $K$, i.e. $3^{2K+1}+2^{K+1}}$ is divisible by $7$.

I think you've meant $3^{2K+1}+2^{K+2}}$...(Nod)
• Aug 21st 2010, 08:53 PM
yuud
Quote:

Originally Posted by Soroban
Hello, yuud!

Here is a sneaky method . . . if you know modulo arithmetic.

$3^{2n+1} + 2^{n+2} \;\;=\;\left(3^3\right)^\frac{2n+1}{3}} + \left(2^3\right)^{\frac{n+2}{3}}$

. . . . . . . . . . . $=\;(27)^{\frac{2n+1}{3}} + (8)^{\frac{n+2}{2}}$

. . . . . . . . . . . $\equiv\; (\text{-}1)^{\frac{2n+1}{3}} + (1)^{\frac{n+2}{2}}\text{ (mod 7)}$

. . . . . . . . . . . $\equiv\;\left(\sqrt[3]{\text{-}1}\right)^{2n+1} + \left(\sqrt[3]{1}\right)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . . $\equiv\; (\text{-}1)^{2n+1} + (1)^{n+2}\text{ (mod 7)}$

. . . . . . . . . . . $\equiv\; -1 + 1\text{ (mod 7)}$

$3^{2n+1} + 2^{2n+1} \;\equiv\;0 \text{ (mod 7)}$

It is divisible by 7 for all positive intergers $n$.

I haven't studied modular arithmetic... but THANKS(Bow)
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