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Find the remainder when 1 1 10 11^{10} is divided by 100.

21 31 03 11 02 71 01 13

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2 solutions

Surya Prakash
Aug 17, 2015

Let 10 C n = C n ^{10}C_{n} = C_{n} 1 1 10 = ( 10 + 1 ) 10 11^{10} = (10+1)^{10} = C 0 1 0 10 + C 1 1 0 9 + + C 9 10 + C 10 = C_{0} 10^{10} +C_{1}10^{9} + \ldots + C_{9} 10 + C_{10} C 9 10 + C 10 101 1 m o d 100 \equiv C_{9} 10 + C_{10} \equiv 101 \equiv \boxed{1} \mod 100

Moderator note:

Good approach of using the binomial theorem and realizing that 1 0 2 = 100 10^2 = 100 .

Another way is to expand ( 100 + 21 ) 5 (100 + 21) ^ 5 , to easily get the factor of 100.

Good one!!

jaiveer shekhawat - 5 years, 10 months ago
Jaiveer Shekhawat
Aug 17, 2015

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