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Find the remainder when \(2^0 + 2^1 + 2^2 + 2^3 + \dots + 2^{100}\) is divided by 7.

Jun 16, 2023

#1
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The remainder when 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^100 is divided by 7 is 0.

This can be found using the following steps:

We can write the sum as a geometric series:

2^0 + 2^1 + 2^2 + 2^3 + ... + 2^100 = 1 + 2 + 4 + 8 + ... + 1024

The sum of a geometric series is equal to a/(1-r), where a is the first term, r is the common ratio, and n is the number of terms. In this case, a=1, r=2, and n=100.

S = a/(1-r) = 1/(1-2) = 1/(-1) = -1

The remainder when a number is divided by 7 is equal to the remainder when the number is divided by 7 and then multiplied by -1. In this case, the remainder when -1 is divided by 7 is 0.

R = (-1)/7 = 0

Therefore, the remainder when 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^100 is divided by 7 is 0.

Jun 16, 2023
#2
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Thank you for the help, but I solved it. The answer is \(3\). Cheers!

Guest Jun 16, 2023
#3
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You lost me at the third line, i.e.,

20 + 21 + 22 + 23 + ... + 2100  =  1 + 2 + 4 + 8 + ... + 1024

2100 is more than 1024, a lot more.  But 210 is 1024.

Is it possible that a zero was dropped during the calculation?