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Neha Mundra
CAT Preparation Community 10y ago
Find the remainder when 10^10+ 10^100+ 101^000 + . . . +10^10000000000 is divided by 7.
3
bindu k 9y ago
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.
Vipul Kumar 7y ago
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
FormulaFree Tutor 5y ago
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
