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if m and n are positive integers, what is the remainder when

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Director
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if m and n are positive integers, what is the remainder when [#permalink] New post 24 Dec 2007, 23:06
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if m and n are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10?

1. n = 2
2. m = 1

What is the short cut without plugging in values for n and solve the exponent?
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SVP
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 [#permalink] New post 24 Dec 2007, 23:28
1. insufficient:
2, sufficient. 3^4n+2 + m = 3^4n+2 + 1
3^4n+2 = 3^2.(3^4)^n is a number ending by 9. So, 3^4n+2 + 1 is a number ending by 0 and therefore, devided by 10
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Re: DS gmatprep remainder, if m and n are positive integers [#permalink] New post 25 Dec 2007, 00:41
gmatnub wrote:
if m and n are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10?

1. n = 2
2. m = 1

What is the short cut without plugging in values for n and solve the exponent?


[9^(2n+1)]/10 + m/10

Irrespective of the value of n, the first term will always have an odd power of 9. Thus we know what remainder will be.
But in second term, we need to know the value of m.

So just by knowing the statement (ii), we can solve this.

Therefore, answer is "B"
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Re: DS gmatprep remainder, if m and n are positive integers [#permalink] New post 25 Dec 2007, 03:08
interesting...

we know that : [3^(4n+2)+m]/10=q + r/10
where q is the quotient and r is the remainder...if we know that
[3^(4n+2)+m] is a multiple of 10, the remainder would clearly be 0.

if n=2 we don't know anything useful; 1 not suff

3^(4n+2) equals 3^4n * 3^2...3^4n is a number which ends in 1 (i.e. 3^4 is 81). if we multiply this number * 3^2 we have a number ending in 9. let's sum with 1...we got 10, so B is suff
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Re: DS gmatprep remainder, if m and n are positive integers [#permalink] New post 22 Jan 2008, 13:54
B
S1 not suffi & not required since the no will end in 9 for all values of n and the remainder depends on m, which is not given in S1.
S2 is suffi since we get the value for m,
Re: DS gmatprep remainder, if m and n are positive integers   [#permalink] 22 Jan 2008, 13:54
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