I came across this example in a video lecture on modular arithmetic. I need to calculate (99^99) modulo 100. It says that since modulo is preserved under multiplication we can just replace 99 in the base by a congruent number that is -1. AFAIK, this is neither modular multiplication or exponentiation.
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3If $a\equiv b\pmod n$ then $a^k\equiv b^k\pmod n$. Indeed, if $a=b+nm$ then $a^k=(b+nm)^k\equiv b^k\pmod n$. – lulu Apr 05 '20 at 11:10
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You can not replace both $99$'s with $-1$, this is only possible for the base. This is simply provable using the fact that
modulo is preserved under multiplication
Let $a \equiv b$ (mod $m$), and we need to prove $a^k \equiv b^k$ (mod $m$). Perform an induction on $k$.
- for $k=1$ this is trivial.
- for $k>1$ we know $a^{k-1} \equiv b^{k-1}$ (mod $m$) by the induction hypothesis, and since modulo is preserved under multiplication, we have $a \cdot a^{k-1} \equiv b \cdot b^{k-1}$, therefore $a^k \equiv b^k$.
For instance $99 \equiv -1$ (mod $100$) implies $99^{99} \equiv -1^{99}$ (mod $100$). which is easy to calculate $-1^{99} = -1$.
Kamyar Mirzavaziri
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Instead of doing $99^{99}$, try this:
$(100-1)^{99}$
Expanding that would bring $100$ terms with a power of $100$ and the last term would be $(-1)^{99}$
That's the real magic of modular arithmetic.
Nεo Pλατo
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