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I have to find the remainder when $5^{99}$ is divided by $13$ without using modular arithmetic. I have to use Binomial Theorem only. Here's my attempt.

My Attempt $$\begin{aligned}5^{99}=125^{33}&=(117+8)^{33}\\ &=\sum_{k=0}^{33}{33\choose k} (117)^{33-k}(8)^{k}\end{aligned}$$

So the answer will be same remainder as obtained when $8^{33}$ is divided by $13$ which would be the same when $5^{11}$ is divided by $13$ which we get by writing $8^{33}$ as $(507+5)^{11}$.

I am not sure on how to proceed from here. Any hints would be appreciated. Thanks

Paras Khosla
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    Isn't this basically modular arithmetic, but disguised? – Rushabh Mehta Dec 12 '19 at 19:13
  • Considering that it is a question about modular arithmetic, I can't imagine what it might mean to solve it without modular arithmetic. – lulu Dec 12 '19 at 19:26
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    @lulu What I meant was that the solution just be build on the most basic concepts (intuitive to someone not familiar with it) of modular arithmetic but the solving be based on binomial theorem. – Paras Khosla Dec 12 '19 at 19:35

3 Answers3

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Hint: $$5^{99} = 5(25)^{49} = 5(26-1)^{49} $$

AgentS
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Hint: $5^2\equiv -1 \bmod 13$

So $5^{98}\equiv (5^2)^{49}\equiv ?? \bmod 13$

Then $5^{99} \equiv ??? \bmod 13$

Keith Backman
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$5^4=625=13\cdot48+1$, and $5^3=13\cdot9+8$.

So $5^{99}=5^{4\cdot24+3}=\left(13\cdot48+1\right)^{24}\cdot\left(13\cdot9+8\right)$.

It follows from the above equation and from the binomial theorem, that the remainder is 8.

user729424
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