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# M025-Question -7

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M025-Question -7 [#permalink]  15 Dec 2009, 19:30
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If $$M$$ and $$N$$ are integers, is $$\frac{10^M + N}{3}$$ an integer?

1. $$N = 5$$
2. $$MN$$ is even

(C) 2008 GMAT Club - m25#7

MN is even in 2nd and N =5 , so that means M can not be 0 or -ve , infact M needs to be even , so MN can be even i.e. M can be 2 , 4 ,6 etc .

now any even value of 10^ M + 5 is divisible by 3 and I am getting C , however that is not mentioned as correct answer , can some one please point out my mistake or solve this question
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Re: M025-Question -7 [#permalink]  15 Dec 2009, 22:13
Even integers can be negative, positive or zero. The set of even integers (OG12 page 108) is given by
{...-4, -2, 0, 2, 4...}
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Re: M025-Question -7 [#permalink]  16 Dec 2009, 12:40
ichha148 wrote:
If $$M$$ and $$N$$ are integers, is $$\frac{10^M + N}{3}$$ an integer?

1. $$N = 5$$
2. $$MN$$ is even

(C) 2008 GMAT Club - m25#7

The question is: (10^m + n)/3= k? where k is an integer.

1. N = 5 is not suff. If m is a +ve integer, k is always an integer.
If m is a -ve integer, k is always a -ve integer. NSF.

2. MN = 2p where p is an integer. In this case, either m or n is even (either -ve or +ve or 0) integer.

If m = 1 and n = 2, k is an integer.
If m = -1 and n = 2, k is not an integer.
If m = 2 and n = 1, k is not an integer.
NSF....

1 and 2: m is even, and n is 5.

When m is -ve even and n = 5, k never be an integer.
When m is +ve even and n = 5, k is always an integer. Still NSF.

E.

ichha148 wrote:
MN is even in 2nd and N =5 , so that means M can not be 0 or -ve , infact M needs to be even , so MN can be even i.e. M can be 2 , 4 ,6 etc .

now any even value of 10^ M + 5 is divisible by 3 and I am getting C , however that is not mentioned as correct answer , can some one please point out my mistake or solve this question

In st 2: Either m or n is even or both can be even. That includes 0 or -ve integers.
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Re: M025-Question -7 [#permalink]  16 Dec 2009, 18:46
Thanks a lot Fremontain and GMAT Tiger, I assumed that even numbers are always positive , a big lesson learnt
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Re: M025-Question -7 [#permalink]  23 Mar 2012, 05:12
1. N does not bring any solution to the rest of the equation so not sufficient.
2.if,MN is even then,
a) M & N both are evens
b) Either of M or N is even
therefore, not sufficient.
3. If we assemble both then, in case, if N=5 i.e odd, then N can be either even or odd , does not signify any answer, so IMO-E
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Re: M025-Question -7 [#permalink]  26 May 2012, 09:55
either of A &B are not sufficient was pretty much clear. solved this problem by pluggin nos. try the following sets for M&N (4,5) & (2,5) and the insufficiency is evident. Tho i totally missed the negative integer point.
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Re: M025-Question -7 [#permalink]  26 May 2012, 15:49
If you use the following sets of values M&N (4,5) & (2,5) , 1&2 will be sufficient.
In fact, for all positive values of M&N , combining 1&2 will be sufficient.
The insufficiency is because M could be negative too, in that case (10^M +N)/3 will not be integer.
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Re: M025-Question -7 [#permalink]  26 May 2012, 20:50
@raingary if you use (4,5) you get 45/3 whihc is an integer by (2,5) gives 25/3 which is not an integer. so these two examples actually illustrate that statement 1&2 are insufficient. This is in addition to the points mentioned about negative values.
Re: M025-Question -7   [#permalink] 26 May 2012, 20:50
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# M025-Question -7

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