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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