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.

Note that you have
$$
7^n - 1 = 6a\\
5^n - 1 = 4b
$$
Now we're interested in whether $a$ and $b$ are even or odd. Which is to say we want to know when $7^n - 1$ is divisible by $4$ (so that when you divide it by $6$ you get an even number), and when $5^n-1$ is divisible by $8$ (so that when you divide it by $4$, you get an even number).
The binomial theorem gives
$$
7^n - 1 = (8-1)^n - 1\\
= 8^n - \binom n18^{n-1} + \cdots + (-1)^{n-1}\binom{n}{n-1}8 + (-1)^n - 1
$$
We see that this is divisible by $4$ exactly when $(-1)^n - 1$ is, which is to say when $n$ is even.
Then we have
$$
5^n - 1 = (4 + 1)^n - 1\\
= 4^n + \binom n14^{n-1} + \cdots + \binom{n}{n-1}4 + 1 - 1
$$
and we see that this is divisible by $8$ precisely when $\binom{n}{n-1} = n$ is even.
So $a$ and $b$ are both even for even $n$, and both odd for odd $n$, proving that $a+b$ is always even, meaning $12(a+b)$ is divisible by $24$.
