# What is the remainder when the expression $\sum_{i=1}^{10} {{10^{10}}^i}$ is divided by 7? [duplicate]

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.

• 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 $k=10^i$ Mar 19, 2023 at 9:29
• Hint: prove by induction $10^{10^i}$ is of the form $7k+4$ for $i\ge1$, using $10^{10^{i+1}}=\left(10^{10^i}\right)^{10}$.
– J.G.
Mar 19, 2023 at 9:41
• Hint : $$10^i = 10k \;where : k \in N$$ Mar 19, 2023 at 9:47
• @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? Jun 2, 2023 at 3:33

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,

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

• Nice but * for multiplication is very computery. \times looks nicer. Mar 19, 2023 at 9:51
• Yes, for multiplication, avoid using * , as it is generally reserved for other things. $\cdot$ and $\times$ (\cdot,\times)are much better choices. Mar 19, 2023 at 10:42
• "for every integer x". Should be "every integer not a multiple of 7"
– D S
Mar 19, 2023 at 11:04
• @K.defaoite i felt it was faster to wright, it's not an official .tex mémoire or anything but thanks i guess
– Zag
Mar 19, 2023 at 13:32
• Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. Mar 19, 2023 at 17:04

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

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