Prove without induction that $2×7^n+3×5^n-5$ is divisible by $24$. I proved this by induction. But I want to show it using modular arithmetic. I tried for sometime as follows
$$2×7^n-2+3×5^n-3\\
2(7^n-1)+3(5^n-1)\\
2×6a+3×4b\\
12(a+b)$$
In this way I just proved that it is divisible by 12 but it is not enough. Am I missing something or it will solved by some other method.

Yes, it can be done by another method. Note that $7^2=2\times24+1$ and that $5^2=24+1$ and that therefore$$7^n\equiv\begin{cases}7\pmod{24}&\text{ if $n$ is odd}\\1\pmod{24}&\text{ otherwise}\end{cases}$$and$$5^n\equiv\begin{cases}5\pmod{24}&\text{ if $n$ is odd}\\1\pmod{24}&\text{ otherwise.}\end{cases}$$So:


*

*if $n$ is odd, then $2\times7^n+3\times5^n-5\equiv2\times7+3\times5-5=24\equiv0\pmod{24}$;

*otherwise, $2\times7^n+3\times5^n-5\equiv2\times1+3\times1-5\equiv0\pmod{24}$.

