Using Induction, prove that $107^n-97^n$ is divisible by $10$ Using Induction, prove that $107^n-97^n$ is divisible by $10$
We need to prove the basis first, so let $ n = 1 $
$107^1-97^1$
$107-97 = 10$ 
This statement is clearly true when $ n = 1 $
Now let's use $P(k)$
$107^k-97^k$ 
So far so good... next I have to use $P(k+1)$ and there is one part that is driving me nuts. The induction hypothesis implies that $107^k-97^k = 10 m$
$107^{k+1}-97^{k+1}$ 
$107^k * 107-97^k *  97$ 
I know that $107^k = 10m+97^k$
$(10m+97^k) * 107 -97^k * 97$
I am lost at this line. What do I do next? Please explain very clearly because I tried to look everywhere online and there aren't very good explanations for after this step. 

Note that 
$$107^{k+1}-97^{k+1}\\=107^k\cdot 107-97^k\cdot 97\\
=107^k\cdot (10+97)-97^k\cdot 97\\
=107^k\cdot 10+107^k\cdot 97-97^k\cdot 97\\
=107^k\cdot 10+(107^k-97^k)\cdot 97$$
where the first term is divisible by $10$, and the second term is also divisible by $10$ by induction. 
