Question :
What is the remainder when
$$ 10^{10} + 10^{10^2} +10^{10^3} + \ldots + 10^{10^{100}} $$
is divided by $7$?
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4Have you heard about Fermat's little theorem? Euler's theorem? – Arthur Jun 22 '15 at 05:25
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Could you please explain ? – Hormigas Jun 22 '15 at 05:26
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1@Anant it's difficult to explain appropriately unless we know which if any of those two theorems sound familiar to you. – Ben Grossmann Jun 22 '15 at 05:35
2 Answers
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While lab's answer is very elegant, it does rely, in some sense, on luck (as do all elegant answers). Here follows a thorough answer that will let you solve any such problem:
First of all, as far as the remainder when divided by $7$ is concerned, there is no difference between $10$ and $3$, so I'm going to work with $$ 3^{10} + 3^{10^2} +3^{10^3} + \cdots + 3^{10^{100}} $$ The remainders when divided by $7$ of successive powers of $3$ goes like this: $$ 3^1 \mapsto 3\\ 3^2 \mapsto 2\\ 3^3 \mapsto 6\\ 3^4 \mapsto 4\\ 3^5 \mapsto 5\\ 3^6 \mapsto 1\\ 3^7 \mapsto 3 $$ and so on. It turns out that the only thing that is important for the $7$-remainder of a power of $3$ is the remainder of the exponent when divided by $6$ (this is what Fermat's little theorem would tell you directly, so you didn't have to check if you knew that one). So we need to find the $6$-remainder of the different powers of $10$.
Now, as for the $6$-remainder, there is no difference between $10$ and $4$, so I will be focusing on the $6$-remainder of $4^n$. Now, the $6$-remainder of the different powers of $4$ are: $$ 4^1 \mapsto 4\\ 4^2 \mapsto 4\\ 4^3 \mapsto 4 $$ and we see that the remainder is the same all the way (Euler's theorem would've told us that the only thing that could matter was whether the exponent was even or not, and we see here that even that doesn't matter).
So we see that the $7$-remainder of $$ 10^{10} + 10^{10^2} +10^{10^3} + \cdots + 10^{10^{100}} $$ is the same as the $7$-remainder of $$ 3^{10} + 3^{10^2} +3^{10^3} + \cdots + 3^{10^{100}} $$ which again is the same as the seven-remainder of $$ 3^{4} + 3^{4^2} +3^{4^3} + \cdots + 3^{4^{100}} $$ which again is the same as the $7$-remainder of $$ 3^{4} + 3^{4} +3^{4} + \cdots + 3^{4} = 100\cdot 3^4 = 8100 $$ and the remainder of $8100$ when divided by $7$ is $1$.
Martin Sleziak
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Arthur
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6
$10^3\equiv-1\pmod7$
For integer $n\ge1,10^n=3\cdot\underbrace{33\cdots33}_{n\text{ digits}}+1$
$\implies10^{10^n}=10(10^3)^{\underbrace{33\cdots33}_{n\text{ digits}}}\equiv10\cdot(-1)^{\underbrace{33\cdots33}_{n\text{ digits}}}\equiv10(-1)\equiv4$ for $n\ge1$
lab bhattacharjee
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