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Tough Remainders Problem

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by tonebeeze » Tue Apr 26, 2011 3:15 pm
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

OA = E

I understood that the key to the problem was to discover the pattern of powers of three, with respect to the units digit:
3^1 = 3
3^2 = 9
3^3 = 7 (27)
3^4 = 1 (81)
3^5 = 3 (243)

but after this point, I lost my way. Can someone please walk me through. Thanks
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by Testluv » Tue Apr 26, 2011 7:05 pm
Whenever there is a rule in the question, we can pick a number that satisfies that rule....n is a positive integer. Let it be 1.

Then, we have 3^11 plus 2. We know that the units digit of successive powers of 3 follow the cycle of 4 you outlined above (other numbers follow other cycles for thier units digit; but they all follow a cycle). So, the units digit of 3^11 is 7. But when we add the 2, the units digit becomes 9. And 9 divided by 5 leaves a remainder of 4.
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by Anurag@Gurome » Tue Apr 26, 2011 8:28 pm
tonebeeze 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

OA = E

I understood that the key to the problem was to discover the pattern of powers of three, with respect to the units digit:
3^1 = 3
3^2 = 9
3^3 = 7 (27)
3^4 = 1 (81)
3^5 = 3 (243)

but after this point, I lost my way. Can someone please walk me through. Thanks
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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Gurome, Inc.
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