What is the remainder when 3^123 is divided by 5?

Question

A. 0
B. 1
C. 2
D. 4
E. 5

gnahanut · Accepted Answer

I believe there is an easier way to do as follow:
3^1 = 3; 3^2 = 9, 3^3 = 7, 3^4 = 1; So basiccally 3 has a cycle of 4.
Therefore 123: 4 = 30 with remainder of 3.
3^3 will have the unit digit of 7, thereby dividing by 5 will leave a remainder of 2.

C.

y4nkee · Answer

Ok lets try this one.

\ 3^{123}= \ (5-2)^{123}= \ 5^{123}+(-2)^{123}

\ 5^{123}  is always divisible by 5.

So need to look at the  \ -2^{123}

Units digit of -2 to the power of n is cyclical and consists of 4 possibilities {-2, 4, -8, 6}

Hence  \ (-2)^{120}  will have last digit 6 and  \ (-2)^{123}  = -8.

Dividing -8 by 5 we have remainder

-8=5n+R
For n=-2
-8=5*-2+R

R=2
Hence Answer is (C)

chetan2u · Answer

hi,
another approach after the highlighted portion 
So need to look at the  \ -2^{123} 
 \ -2^{123}= (-2^4)^{\frac{123}{4}} = 16^{30}*-2^3=(15+1)^{30}*-8 
now (15+1)^{30}  will leave 1 as remainder, so total remainder = 1*-8 = -8..
as correctly done, we cannot have negative remainder so remainder = 10-8 = 2

y4nkee · Answer

Simple and straight to the point!

Sidart · Answer

Q. 3^123 % 5

= (5-2)^123 % 5
 * all the factors will be divisible by 5 except (-2)
= (-2)^123%5
 * taking in account the cyclicity of 2; ie 4
= (-2^4)^30*(-2)^3 % 5
= 16^30*(-16) % 5
 * 16%5 leaves remainder as 1 
= (1)^30*(-16) % 5
= (-16) % 5 leaves remainder as -1; ie 5 + (-1) = 4

BrushMyQuant · Answer

We know to find what is the remainder when  3^{243}  is divided by 5

Theory: Remainder of a number by 5 is same as the unit's digit of the number

( Watch this Video  to Learn  How to find Remainders of Numbers by 5 )

Using Above theory , Let's find the unit's digit of  3^{243}  first.

We can do this by finding the pattern / cycle of unit's digit of power of 3 and then generalizing it.

Unit's digit of  3^1  = 3
Unit's digit of  3^2  = 9
Unit's digit of  3^3  = 7
Unit's digit of  3^4  = 1
Unit's digit of  3^5  = 3

So, unit's digit of power of 3 repeats after every  4^{th}  number.
=> We need to divided 243 by 4 and check what is the remainder
=> 243 divided by 4 gives 3 remainder

=>  3^{243}  will have the same unit's digit as  3^3  = 7
=> Unit's digits of  3^{243}  = 7

But remainder of  3^{243}  by 5 cannot be more than 5
=> Remainder = Remainder of 7 by 5 = 2

So,  Answer will be C 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

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What is the remainder when 3^123 is divided by 5?

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