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by diehard_gmat » Fri Mar 18, 2011 12:19 am
If n is a positive integer, what is the remainder when [3^(8n+3) + 2] is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4
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by Anurag@Gurome » Fri Mar 18, 2011 1:00 am
diehard_gmat wrote:If n is a positive integer, what is the remainder when [3^(8n+3) + 2] is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4
The units digits of the powers of 3 has a cycle of 3, 9, 7, 1, 3, 9, ...

Hence, units digit of 3^(Some multiple of 4 + 3) = units digit of 3^3 = units digit of 27 = 7

Hence, units digit of [3^(8n + 3) + 2] = 7 + 2 = 9

Hence, required remainder = 9 - 5 = 4

The correct answer is E.
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by manpsingh87 » Fri Mar 18, 2011 1:07 am
diehard_gmat wrote:If n is a positive integer, what is the remainder when [3^(8n+3) + 2] is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4
unit digit of 3 repeats itself after every 4 powers for example, 3^1=3;3^2=9;3^3=7;3^4=1;3^5=3

now when we divide (8n+3) by 4 remainder will be 3 i.e. unit digit will always be 7;
therefore unit digit of 3^(8n+3) + 2 will be 9;
hence when divided by 5 remainder will be 4!! option E
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