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Find the remainder when the sum 2^{2} + 2^{6} + 2^{10} + 2^{14} + \cdots + 2^{110} + 2^{114} + 2^{118} is divided by 10.

 Jul 6, 2022
 #1
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0

sumfor(n, 1, 30, 2^(4*n - 2))==354460 7988759775 6610768188 2741425220 mod 10 ==0 - the remainder.

 Jul 6, 2022
 #2
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2^2 mod 10 = 4

2^6 mod 10 = 4

2^10 mod 10 = 4

any 2^(4k+2) mod 10 = 4, k≥0, can you see why?

(118-2)/4 + 1 = 30

30 * 4 mod 10 = 120 mod 10 = 0 mod 10.

answer is 0


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