Question

Show that ${4.6}^n+5^{n+1}$ when divided by $20$ leaves remainder $9$.

Answer 1

Let $f\left(n\right)=4.6^n+5^{n+1}-9$

$f\left(1\right)=4.6^1+5^{1+1}-9=40$

which is divisible by $20$

$f(n+1)=4.6^{n+1}+5^{n+2}-9f(n+1)-f\left(n\right)=4.6^{n+1}+5^{n+2}-9-4.6^n-5^{n+1}+9f(n+1)-f\left(n\right)=4.6^{n+1}-4.6^n+5^{n+2}-5^{n+1}f(n+1)-f\left(n\right)=4.6^n(6-1)+5^{n+1}(5-1)f(n+1)-f\left(n\right)=20.6^n+20.5^{n}f(n+1)-f\left(n\right)=20(6^n+5^n)$

which is a multiple of $20$

We subtracted $9$ from the given equation which become divisible by $20$

So the given equation leaves remainder $9$ when divided by $20$