<s>[INST] If $n\geq 3$, $9^n \equiv a (\mod 100)$ and $9^{n+1} \equiv b (\mod 100)$, then $a+b=90$. I noticed a pattern in the powers of 9 modulo 100.
$9^1 \equiv 9 \pmod{100}$
$9^2 \equiv 81 \pmod{100}$
$9^3 \equiv 29 \pmod{100}$
$9^4 \equiv 61 \pmod{100}$
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and conjectured the following:
If $n\geq 3$ is an odd integer where $9^n \equiv a \pmod{100}$ and $9^{n+1} \equiv  b \pmod{100}$, then $a+b=90$.
How do I prove this?
 [/INST] $\ \overbrace{ \color{#c00}{10}\cdot 9^n}^{\textstyle 9^n\!+\!9\cdot9^n\!\!\!\!\!\!\!\!}\bmod \color{#c00}{10}0\, =\, \color{#c00}{10}\,(\!\!\!\overbrace{9^n}^{\large  (-1)^{\Large n}}\!\!\!\bmod 10)\, = \,\left\{\begin{align}&10,\,\ n\rm\ even\\ &90,\,\ n\ \rm odd\end{align}\right.$
$\text{using}\,\ \color{#c00}ab\bmod \color{#c00}ac\, =\, \color{#c00}a\,(b\bmod c)\, = $ $\bmod\!$ Distributive Law to factor out $\,\color{#c00}{a = 10}$
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