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

Question

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 $k=10^i$ — Henry, Mar 19 '23 at 09:29
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 '23 at 09:41
@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? — Apass.Jack, Jun 02 '23 at 03:33

Zag · Accepted Answer · 2023-03-19T13:34:17.987

3

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,

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

edited Mar 19 '23 at 13:34

answered Mar 19 '23 at 09:44

Zag

977

Nice but * for multiplication is very computery. \times looks nicer. – badjohn Mar 19 '23 at 09:51
Yes, for multiplication, avoid using * , as it is generally reserved for other things. $\cdot$ and $\times$ (\cdot,\times)are much better choices. – K.defaoite Mar 19 '23 at 10:42
"for every integer x". Should be "every integer not a multiple of 7" – D S Mar 19 '23 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 '23 at 13:32
@DS you are absolutely right ! – Zag Mar 19 '23 at 13:34
1

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 19 '23 at 17:04

score 1 · Answer 2 · answered Mar 19 '23 at 10:35

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. — Bill Dubuque, Mar 19 '23 at 17:05

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

2 Answers2