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

Question

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 · Accepted Answer

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

Kody · Answer

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.

Green12345554 · Answer

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

lacktutor · Answer

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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repellatquas · Answer

Another method:
5^37 mod 63 = 5^(3*12+1) mod 63
=   mod 63
= (( 5^3)^12) mod 63 * 5 mod 63
= -1^12 * 5 mod 63
= 1 * 5  
= 5
ANS E

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