If n and m are positive integers, what is the remainder when : Quant Question Archive [LOCKED]
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If n and m are positive integers, what is the remainder when

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If n and m are positive integers, what is the remainder when [#permalink]

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28 Jan 2008, 15:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10?

1. n = 2
2. m = 1
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28 Jan 2008, 15:27
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3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
3^6=729
etc

and there's a pattern.

3^(4n+2) means that for whatever integer n it will be 3^6, 3^10, 3^14, etc...and since we're dealing with a cycle of 4 (see above) they'll all have the same remainder when divided by 10. In this case that remainder will be 9. We just need to know what m is. Statement 2 is sufficient, and there will be no remainder.

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29 Jan 2008, 19:16
I don't see how the remainder is always 9. Can you explain this?
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29 Jan 2008, 19:23
jimmyjamesdonkey wrote:
I don't see how the remainder is always 9. Can you explain this?

In 3^(4n+2), substitute n = 0, 1 etc.
when n = 0, 3^(4n+2) = 3^2 = 9
when n = 1, 3^(4n+2) = 3^6 = 729
.
.
and so on

The remainder of these numbers, when divided by 10, is always 9.

Hence its independent of the first term, namely 3^(4n+2). What matters is the value of the second term, m.
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29 Jan 2008, 19:31
Yeah I get that...but take 81/10 = 8.1. Where is the remainder 9 there? It's only r1.
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29 Jan 2008, 19:40
jimmyjamesdonkey wrote:
Yeah I get that...but take 81/10 = 8.1. Where is the remainder 9 there? It's only r1.

For what value of n, did you get 81?
3^(4n + 2) = 81
or, 4n + 2 = 4
n = 0.5 (invalid)

4n+2 returns 2, 6, 10 etc.. and every 4th number in this sequence has 9 in the unit's position
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29 Jan 2008, 19:47
Good point, I was reading the 2nd post:

3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
3^6=729
etc
Re: GMATprep : remainder   [#permalink] 29 Jan 2008, 19:47
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