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What is the remainder when 5^37 is divided by 63 ?

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Bunuel
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What is the remainder when 5^37 is divided by 63 ? [#permalink] New post   24 Feb 2020, 03:55
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A
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E

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56% (01:35) correct 44% (01:49) wrong based on 161 sessions
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What is the remainder when \(5^{37}\) is divided by 63 ?

A. 1
B. 2
C. 3
D. 4
E. 5

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lacktutor
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Re: What is the remainder when 5^37 is divided by 63 ? [#permalink] New post   24 Feb 2020, 05:13
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What is the remainder when \(5^{37}\) is divided by 63 ?

\(5^{37}= 5*5^{36}= 5*125^{12}= 5(126-1)^{12}= 5(63*2-1)^{12}= 5*(63m+...+1)\\
\)

--> Dividing by 63, the remainder will be 5.

The answer choice E is correct
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Kody
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Re: What is the remainder when 5^37 is divided by 63 ? [#permalink] New post   24 Feb 2020, 05:16
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5^37 = 5^36 * 5
5^36 = 125^12
(125^12)/63 has a remainder of -1
-1^12 = 1
1 * 5(from the first line of the equation) = 5
Answer is E.
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Green12345554
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Re: What is the remainder when 5^37 is divided by 63 ? [#permalink] New post   26 Feb 2020, 21:04
can you please explain why this doesn't work --->

5^37/63 = 125^34/63 = (63+62)^34/63

Using binomial theorem:
(63m+ .. + 62^34 )/63 = 62^34/63

(63-1)^34/63 = (-1)^34/ 63 = 1
i.e. option A
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Re: What is the remainder when 5^37 is divided by 63 ? [#permalink] New post   26 Feb 2020, 22:13
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Green12345554 wrote:
can you please explain why this doesn't work --->

5^37/63 = 125^34/63 = (63+62)^34/63

Using binomial theorem:
(63m+ .. + 62^34 )/63 = 62^34/63

(63-1)^34/63 = (-1)^34/ 63 = 1
i.e. option A


hi,
The highlighted part above is not correct. It should be
—> \(5^{37} = 5^3*5^{34}\) or

—> \(5^{37} = 5*5^{36}= 5*5^{3*12} = 5*125^{12}\)

Hope it helps

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Re: What is the remainder when 5^37 is divided by 63 ? [#permalink] New post   25 Mar 2022, 02:51
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Re: What is the remainder when 5^37 is divided by 63 ? [#permalink]
25 Mar 2022, 02:51
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