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What is the remained when $\sum_{i=1}^{10} {{10^{10}}^i} $ is divided by 7? I have tried writing $10 = 3+7$ which makes the expression $\sum_{i=1}^{10} {{(3+7)^{10}}^i}= \sum_{i=1}^{10} {(7p+{{(3)^{10}}^i})} $ where p is an integer. Therefore we get the same remainder as when $\sum_{i=1}^{10} {{{3^{10}}^i}}$ is divided by 7. Please mention how to simplify further.

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    $\begingroup$ The next step is to consider the remainder when $3^k$ is divided by $7$ (try some small $k$ and spot the repeating pattern), and what this means when $k=10^i$ $\endgroup$
    – Henry
    Mar 19, 2023 at 9:29
  • $\begingroup$ Hint: prove by induction $10^{10^i}$ is of the form $7k+4$ for $i\ge1$, using $10^{10^{i+1}}=\left(10^{10^i}\right)^{10}$. $\endgroup$
    – J.G.
    Mar 19, 2023 at 9:41
  • $\begingroup$ Hint : $$10^i = 10k \;where : k \in N $$ $\endgroup$ Mar 19, 2023 at 9:47
  • $\begingroup$ @Suprativ It looks you come from a country where people cannot vote. You have accepted one answer but you don't believe it is helpful to you, which sounds like you accepted a governor without voting. Why not upvote? $\endgroup$
    – Apass.Jack
    Jun 2, 2023 at 3:33

2 Answers 2

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Thanks to Fermat's little theorem, $x^6$ has remainder $1$ when divided by $7$, for every integer $x$ not divisible by $7$. So for every $i = 1, \ldots, 10$, $10^i = (6+4)^i$, hence $10^i$ has remainder $4^i$ when divided by $6$. Thus, $3^{10^i} \equiv 3^{4^i}$ ($\equiv$ means "has the same remainder when divided by $7$).

But $4\times 4 = 16$ which has remainder $4$ when divided by $6$, so $4^i$ has remainder $4$ when divided by $6$. Hence, $3^{10^i} \equiv 3^4 = 81$, which has remainder $4$ when divided by $7$. Hence,

$$\sum_{i=1}^{10} 10^{10^i} \equiv 10\times 4 \equiv 5.$$

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  • $\begingroup$ Nice but * for multiplication is very computery. \times looks nicer. $\endgroup$
    – badjohn
    Mar 19, 2023 at 9:51
  • $\begingroup$ Yes, for multiplication, avoid using * , as it is generally reserved for other things. $\cdot$ and $\times$ (\cdot,\times)are much better choices. $\endgroup$
    – K.defaoite
    Mar 19, 2023 at 10:42
  • $\begingroup$ "for every integer x". Should be "every integer not a multiple of 7" $\endgroup$
    – D S
    Mar 19, 2023 at 11:04
  • $\begingroup$ @K.defaoite i felt it was faster to wright, it's not an official .tex mémoire or anything but thanks i guess $\endgroup$
    – Zag
    Mar 19, 2023 at 13:32
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    $\begingroup$ Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. $\endgroup$ Mar 19, 2023 at 17:04
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answers : $$10^i=10k\;\;k\in\;N\\~\\10^i≡1[3]\;\implies\;10k≡1[3]\;\implies\;k=3k′+1\;k′\;\in\;N\\~\\\\~\\10^5≡5[7]\implies\;10^{10k}≡5^{2k}[7]\implies\;10^{10k}≡5^{2(3k′+1)}[7]\implies\;10^{10k}≡5^{6k′+2}[7]\\~\\\forall\;k′\;\in\;N\;;5^{6k′+2}≡4[7]\\~\\\implies10^{10k}≡4[7]\\~\\\implies\;10^{10^i}≡4[7]\\~\\\sum_{i=1}^{10}10^{10^i}≡(\sum_{i=1}^{10}4)[7]\implies\sum_{i=1}^{10}10^{10^i}≡5[7]$$

the remainder Is : 5

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    $\begingroup$ Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. $\endgroup$ Mar 19, 2023 at 17:05