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number properties

Expert replies
Source: — Data Sufficiency |

by neelgandham » Fri Jan 20, 2012 7:12 am
m and n are integers, is m^n an integer?
1) n^m is positive
If m=2 and n=-1
n^m = (-1)^2(=1) is positive and m^n = (2)^-1 = 1/2 is not an integer
If m=2 and n=3
n^m = (3)^2(=9) is positive and m^n = (2)^3 = 8 is an integer.
Insufficient to answer the question
2) n^m is an integer
n^m = (-1)^2(=1) is an integer and m^n = (2)^-1 = 1/2 is not an integer
If m=2 and n=3
n^m = (3)^2(=9) is an integer and m^n = (2)^3 = 8 is an integer.
Insufficient to answer the question
From 1 and 2
n^m = (-1)^2(=1) is a positive integer and m^n = (2)^-1 = 1/2 is not an integer
If m=2 and n=3
n^m = (3)^2(=9) is a positive integer and m^n = (2)^3 = 8 is an integer.
Insufficient to answer the question

IMO E=
Anil Gandham
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by tpr-becky » Fri Jan 20, 2012 11:11 am
First issue is to discover what they are testing - if both nubmers are integers then teh only way form m^n to be a non-integer if n is negative (thus creating a reciprocal). You want to look at each of the statements to determine whether either one forces n to be negatve .

1) says that either n is positive or m is an even integer (not what we are looking for so insufficient)

2) says that m is positive (doesn't create a reciprocal) but no information about n as n could be positive or negative.

if we put them both together we learn that m is positive and might be even and than n might be positive - no info we need therefore the answer is E.
Becky
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The Princeton Review
Irvine, CA
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by pinchharmonic » Fri Jan 20, 2012 12:10 pm
tpr-becky wrote:First issue is to discover what they are testing - if both nubmers are integers then teh only way form m^n to be a non-integer if n is negative (thus creating a reciprocal). You want to look at each of the statements to determine whether either one forces n to be negatve .

1) says that either n is positive or m is an even integer (not what we are looking for so insufficient)

2) says that m is positive (doesn't create a reciprocal) but no information about n as n could be positive or negative.

if we put them both together we learn that m is positive and might be even and than n might be positive - no info we need therefore the answer is E.
Becky,

I have a question about my approach. Taking the statement, I concluded that if M^N is an integer I have two cases:

N Positive OR M = 1
i don't consciously think of the opposite condition, but i would write it down just in case:
N negative AND M not equal 1

then for N^M being positive

N positive allows M to be anything
N negative requires M to be even

N is not forced positive, and M is not forced to be 1 in the cases where N is negative. insuff

then for N^M being an integer

N can be anything, M positive
N must be 1, M negative

N is not forced positive, and when N is negative M is not forced to be 1 either. insuff

then combining two

N positive, M positive
N negative, M even

again N is not forced positive, and when N is negative M can't even be 1. insuff.

etc.

since i'm looking for either of two things to occur (N and M, instead of just N like every other person who solved it did), it gets very confusing and ends up taking way too much time. it seems easier to just take some examples and plug them in and solve it quickly.

but that makes me wonder about hard problems. This problem has so many opportunities to become difficult in the details... but if it's a straightforward problem you end up wasting so much time doing it in detail.


for example, say you are given a statement that leads you with the below conclusions. You don't even bother to figure out what M could be since N is already positive and negative.

N can be positive
N can be negative

then instantly most people would mark that as insufficient, since N is not fixed + or -. But what if you went further and realized

N can be positive, M can be anything
N can be negative, M must be 1.

then in that case A proves the statement either N being positive OR M being 1 is sufficient.

and another possibility would be:

N must be negative
then you might be inclined to think, ok, M^N is never an integer. but what if M was 1?
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by ArunangsuSahu » Sat Jan 21, 2012 1:50 am
Check these things
1) If Statement 1 is making sure that N can not be -ive integer
2) If Statement 2 is making sure that N can not be -ive integer
3) Combining also if N is +ive
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