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.

Case 1 : $n$ is odd
In this case $$2×7^n+3×5^n-5{=2×7^n+2×5^n+5^n-5\\=2\times\underbrace{(7^n+5^n)}_{12k}+5(5^{n-1}-1)\\=24k+5(\underbrace{25^{n-1\over 2}-1}_{24k'})\\=24k''}$$
Case 2 :  $n$ is even 
In this case $$2×7^n+3×5^n-5{=14×7^{n-1}+15×5^{n-1}-5\\=14\times\underbrace{(7^{n-1}+5^{n-1})}_{12k}+5(5^{n-2}-1)\\=24k+5(\underbrace{25^{n-2\over 2}-1}_{24k'})\\=24k''}$$
