M025-Question -7

Question

If  M  and  N  are integers, is  \frac{10^M + N}{3}  an integer?

1.  N = 5 
 2.  MN  is even

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

Fremontian · Answer

Even integers can be negative, positive or zero. The set of even integers (OG12 page 108) is given by
{...-4, -2, 0, 2, 4...}

GMAT TIGER · Answer

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.

In st 2: Either m or n is even or both can be even. That includes 0 or -ve integers.

ichha148 · Answer

Thanks a lot Fremontain and GMAT Tiger, I assumed that even numbers are always positive , a big lesson learnt

thevenus · Answer

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

vibhav · Answer

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.

raingary · Answer

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.

vibhav · Answer

@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.

M025-Question -7

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