mindmind wrote:

Find the remainder when the sum of \(2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}\) is divided by 7

A. 0

B. 2

C. 1

D. 6

E. 5

Solved this in one minute.

Just remember to use the concept explained @

quarter-wit-quarter-wisdom-a-remainders-post-for-the-geek-in-you and half of the remainder Qs become a piece of cake

All the terms in the expression (a+b)^z are divisible by a except b^z(8=7+1) Make the modifications in the expression so that it is in some relation with the given divisor (usually plus or minus 1)

2^100 = 2* 8^33 = 2* (7+1)^33

2^200 = 4* 8^66 = 4* (7+1)^66

2^300 = 1* 8^100 = 1* (7+1)^100

2^400 = 2* 8^133 = 2* (7+1)^133

2^500 = 4* 8^166 = 4* (7+1)^166

2^600 = 1* 8^200 = 1* (7+1)^200

Observer that this it is a repeat cycle of 2, 4 and 1

Make a note of total number of terms = 100

remainder from first term = 2*1

remainder from second term = 4*1

remainder from third term = 1*1

Sum of first three remainders = 7. Remainder =0 .Sum of first 99 remainders = 7*33 Remainder =0 . Remainder on the 100th term = 2 (from the cycle)

Answer B

Though I sound pretty confident I need someone to validate the method

Thanks!

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