<s>[INST] Best method for proving that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$ I am asked to prove by induction that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$.
I wonder whether there is a more direct method, for example factorizing by $10$.
If an expression is divisible by $10$, does this mean that I can factorize it by $10$?
Thanks in advance
 [/INST] Best method?  Depends on what you know.
If you know modular arithmetic and Eulers Theorem then
$7\cdot 11^{2n+1} - 3^{4n-1}\equiv 7\cdot 1^{2n+1} - (3^{-1})\cdot 3^{4n}\equiv 7-(3^{-1})\pmod{10}$ and as $3\cdot 7 \equiv 1 \pmod {10}$ then $3^{-1}\equiv 7 \pmod {10}$ and $7-7\equiv 0\pmod {10}$ so $10|7\cdot 11^{2n+1} - 3^{4n-1}$.
But that assumes you are comfortable with many concepts
If you don't know any modular arithmetic:
$11^k = (10+1)^k = 10^k + k10^{k-1} + ...... + k\cdot 10 + 1 = 10 M + 1$.
And $3^{4n-1} = 3\times 3^{4n-2}= 3\times 9^{2n-1}$.  And if $k=2n-1$ is odd the $9^k = (10-1)^k = 10^k - k10^{k-1} + .......   + k\cdot 10 - 1= 10N-1$.
So $7\cdot 11^{something} - 3^{1+2\times something\ odd} = 7(10M + 1)- 3(10N-1)= 70M -30N +10$.
</s>