Find the remainder when 10^10+ 10^100+ 101^000 + . . . +10^10000000000 is divided by 7.

All you need is basic modular arithmetic. No totient rule needed.

Since 10 is congruent to 3 (Mod 7), 10^10 is congruent to 3^10 (mod 7).

3^10 = 9^5, which is congruent to 2^5 = 32 (mod 7). When we divide 32 by 7, we get a remainder of 4. Therefore, 10^10 is congruent to 4 (mod 7).

In order to find what 10^100 is congruent to mod 7, we raise both sides of the previous equation by the power of 10 to get

(10^10)^10 = 4^10 (mod 7).

4^10 = 16^5 (mod 7), which is congruent to 2^5 (mod 7). We know that 10^100 is congruent to 4 mod 7.

Repeating this process of taking the power of 10, we discover that each term in the series has remainder 4 mod 7.

There are 10 terms in the series, which adds up to a sum of 10*4 = 40, which is congruent to 5 mod 7.

ANSWER: 5

first we reduce it to : (3^10+3^100+....+3^10000000000)/7 

Apply totient rule each term, -> 3^4+3^4+....+3^4(10 terms)/7 [totient of 7=6 and remainder of power of 10 divided by 6 is always 4]

=remainder(81*10/7)= (4*3)/7= 5 answer

Hi Neha,

This question can be solved by using cyclicity concept.

10^1/7 ---> remainder 3

10^2/7 ---> remainder 2

10^3/7 ---> remainder 6

10^4/7 ---> remainder 4

10^5/7 ---> remainder 5

10^6/7 ---> remainder 1

And the remainder cycle repeats.

So, 10^10 leaves remainder 4

10^100 - 5

10^1000 - 1

and so on..

10^10000000000 - 3

Sum of all these remainders = 4+5+1+3+2+6+4+5+1+3=34

Remainder when 34 divided by 7 is 6.

Hope this helps

Thanks.