# Thread: multiple of 5

1. ## multiple of 5

Prove that, for any positive integer n, 3^3n+1 + 2^n+1 is always divisible by 5.

2. Originally Posted by bearej50
Prove that, for any positive integer n, 3^3n+1 + 2^n+1 is always divisible by 5.
For n= 1, that says $3^{1}+ 2^{1}= 5$ is divisible my 5 which is true.

Assume that, for some positive integer k, $3^{3k+1}+ 2^{k+1}$ is a multiple of 5. That is, that $3^{3k+1}+ 2^{k+1}= 5i$ for some integer i.

$3^{3(k+1)+1}+ 2^{(k+1)+1}= (3^{3k+ 4}+ 2^{(k+1)+ 1}= 3^{3}3^{3k+1}+ 2(2^{k+1}$
"Add and subtract" $3^3(2^{k+1}$ to get
$= 3^3(3^{k+1}+ 2^{k+1})- 3^3(2^{k+1})+ 2(2^{k+1}$
$= 27(3^{k+1}+ 2^{k+1})- (27- 2)2^{k+1}$
$= 27(5i)- (25)2^{k+1}= 5(27i- (5)2^{k+1}$
which is 5 times an integer.

3. thank you so much

4. Hello, bearej50!

Prove that, for any positive integer $n\!:\;\;3^{3n+1} +2^{n+1}$ is always divisible by 5.

We have: . $N \;=\;3^{3n}\cdot3^1 + 2^{n+1} \;=\;(3^3)^n\cdot3 + 2^{n+1} \;=\;3(27^n) + 2^{n+1} \;= \;3(25 + 2)^n + 2^{n+1}$

. . $= \;3\bigg[25^n + {n\choose1}25^{n-1}\cdot2 + {n\choose2}25^{n-2}\cdot 2^2 + \hdots + {n\choose 2}25^2\cdot2^{n-2} + {n\choose1}25\cdot2^{n-1} + 2^n\bigg] + 2^{n+1}$

. . $= \;3\underbrace{\bigg[25^n + {n\choose1}25^{n-1}\cdot 2 + \hdots + {n\choose1}25\cdot2^{n-1}\bigg]}_{\text{This is a multile of 5, }5k} \;+\; 3\cdot2^n \;+\; 2^{n+1} \;= \;5k \;+\; 3\cdot2^n \;+\; 2^{n+1}$

$\text{Factor: }\;N \;=\; 5k + (3 + 2)\!\cdot\!2^n \;=\;5k + 5\!\cdot\!2^n \;=\;\underbrace{5(k + 2^n)}_{\text{multiple of 5}}$

Therefore, $N$ is divisible by 5.

5. thank you