Prove by induction that $99 | 10^{2n} + 197$ for $n\ge 1$ I'm not sure whether I should make use of the transitive property, or this $a|b\Rightarrow b = a*z$ / $z\in\mathbb Z$ to solve the problem.
I'm mainly looking to solve it through induction using the properties of the integer numbers.
 A: It is tagged "induction" so we induct: $ $let  $\,f(n)= 100^n+197.\,$ Then $\,99\mid f(0)=198.\,$
Also $\,\ 99\mid \color{#c00}{f(n\!+\!1)-f(n)} =100^{n+1}-100^n = (100-1)100^n  = 99\cdot 100^n.\,$
Thus $\,99\mid \color{#0a0}{f(n)}\,\Rightarrow\,99\mid\,f(n\!+\!1)=\color{#c00}{f(n+1)-f(n)}+\color{#0a0}{f(n)},\,$ completing the induction.
A: We have
$$10^2\equiv 1 \mod 99$$
so
$$10^{2n}+197\equiv 1+197=198\equiv0\mod 99$$
A: $$10^{2n} + 197$$
$$=(99+1)^n + 198 -1$$
Using binomial expansion, we can say all the terms in $(99+1)^n$ are multiples of $99$ except the last term $nC_n*1^{99}$ which is $1$.
So:$$(99+1)^n = 99k + 1$$
$$10^{2n} + 197 = 99k + 1 + 198 -1 = 99(k+2)$$
This is clearly divisible by $99$.
Using Induction:
$$10^{2(n+1)} + 197 = 100*100^n+197$$
From inductive assumption, we already know that: $$100^n+197 = 99k$$
$$100^n = 99k-197 = 99(k-2)+1 = 99l+1$$
Substituting this in the first equation:
$$100*(99l+1)+197 = 9900l+100+197 = 9900l+99*3 = 99(100l+3)$$
Hence proved by induction as well !
