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24 Feb 2020, 05:30
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B
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Difficulty:
85%
(hard)
Question Stats:
45% (01:59) correct
55%
(02:18)
wrong
based on 208
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What is the remainder when \(52^{60}\) is divided by 31 ?
A. 0
B. 1
C. 21
D. 23
E. 30
Are You Up For the Challenge: 700 Level Questions
A. 0
B. 1
C. 21
D. 23
E. 30
Are You Up For the Challenge: 700 Level Questions
24 Feb 2020, 07:35
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Bunuel
Working out a pattern will be cumbersome, so let us do by BINOMIAL theorem..
\(52^{60}=(62-10)^{60}\), so remainder will be \((-10)^{60}=100^{30}=(93+7)^{30}\)..
Remainder when \((93+7)^{30}\) is divided by 31 will be \(7^{30}=(7^3)^{10}=343^{10}=(341+2)^{10}=(31*11+2)^{10}\)
Remainder when \((31*11+2)^{10}\) is divided by 31 will be\( 2^{10}=(2^5)^2=32^2=(31+1)^2\)
Thus the remainder = \(1^2\) or 1
B
General Discussion
29 Mar 2020, 03:11
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Bunuel
52 = 62 - 10
52^60mod31 = (-10)^60mod31 = 10^60mod31 = 7^30mod31 = 2^10mod31 = 1^2mod31 = 1
IMO B
