b) Ta có: M = 3n + 2 + 2n + 3 + 3n + 2n + 1 = 3n . 32 + 3n + 2n . 23 + 2n . 2 = 3n . (32 + 1) + 2n . (23 + 2) = 3n . 10 + 2n . 10 = (3n + 2n) . 10 \(⋮\) 10.
Vậy...
a: n chia5 dư 1
n chia 8 dư 4
Do đó: \(\left\{{}\begin{matrix}n-1\in\left\{0;5;10;...\right\}\\n-4\in\left\{0;8;16;....\right\}\end{matrix}\right.\Leftrightarrow n=41\)
b: \(M=3^n\left(3^2+1\right)+2^n\left(2^3+2^2\right)=3^n\cdot10+2^n\cdot10⋮10\)