If $n\geq 3$, $9^n \equiv a (\mod 100)$ and $9^{n+1} \equiv b (\mod 100)$, then $a+b=90$.

Question

I noticed a pattern in the powers of 9 modulo 100.

$9^1 \equiv 9 \pmod{100}$

$9^2 \equiv 81 \pmod{100}$

$9^3 \equiv 29 \pmod{100}$

$9^4 \equiv 61 \pmod{100}$

. . .

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?

Answer 1

$$a+b\equiv9^n(1+9)\pmod{100}$$

Now $9^n\equiv(-1)^n\pmod{10}\implies10\cdot9^n\equiv10\cdot(-1)^n\pmod{10\cdot10}$

Answer 2

From the binomial theorem, it follows that,

for $n$ odd, $9^n=(10-1)^n\equiv10n-1\bmod100$,

and $9^{n+1}=(10-1)^{n+1}\equiv-10(n+1)+1\bmod100$.

Therefore, for $n$ odd, $9^n+9^{n+1}\equiv-10\equiv90\bmod100$.

Answer 3

$\ \overbrace{ \color{#c00}{10}\cdot 9^n}^{\textstyle 9^n\!+\!9\cdot9^n\!\!\!\!\!\!\!\!}\bmod \color{#c00}{10}0\, =\, \color{#c00}{10}\,(\!\!\!\overbrace{9^n}^{\large (-1)^{\Large n}}\!\!\!\bmod 10)\, = \,\left\{\begin{align}&10,\,\ n\rm\ even\\ &90,\,\ n\ \rm odd\end{align}\right.$

$\text{using}\,\ \color{#c00}ab\bmod \color{#c00}ac\, =\, \color{#c00}a\,(b\bmod c)\, = $ $\bmod\!$ Distributive Law to factor out $\,\color{#c00}{a = 10}$