If m and n are positive integers, is root (n-m) an integer?

Question

If  m  and  n  are positive integers, is  \sqrt{n-m}  an integer?

(1)  n > m + 15 
(2)  n = m(m+1)

KUDO please, if you like it.

ENGRTOMBA2018 · Accepted Answer

m,n are positive integers and we need to find whether  \sqrt{n-m} = integer .

Per statement 1, n>m+15.

Take 2 cases:

case 1, when n=m+16, then you get "yes" for  \sqrt{n-m} = integer  but
case 2, when n=m+17, then you get "no" for  \sqrt{n-m} = integer

Thus this statement is not sufficient.

Per statement 2, n=m(m+1) ---> \sqrt{n-m} = \sqrt{m^2} = m =integer. Sufficient.

B is the correct answer.

vaibhav1221 · Answer

Is  \sqrt{n - m}  an integer, given that  m  and  n  are positive integers?

Statement 1

n > m + 15

Easiest way to say that this statement is insufficient would be by thinking of it as statement that says,  n > m 
This just means that  \sqrt{n - m}  is not an irrational number. This is insufficient information to answer the question

Another point of view is as follows
Assume that  \sqrt{n - m}  is a positive integer greater than another positive integer, say x.

\sqrt{n - m} > x 
squaring both sides
 n - m = x^2 
 n = m + x^2

This equation is similar to statement 1, however, one may be tempted to say that  \sqrt{15}  is not an integer, hence, the whole argument falls apart and  \sqrt{n - m}  is not an integer. But that's not the case.

If  n > m + 15 
then,  n  is also greater than  m + 9 
and  \sqrt{9}  is an integer.

Also, if  n > m + 15 
then,  n  is also greater than  m + 4  and  n  is also greater than  m + 1 
and  \sqrt{4} ,  \sqrt{1} are integers.

Two different answers, hence the statement is insufficient.

Statement 2

n = m(m+1) 
 n = m^2+m ....... (#1)

Substituting (#1) in  \sqrt{n - m}

\sqrt{m^2 + m - m} 
 \sqrt{m^2} 
 m

m is an integer, as stated in the question, hence,  \sqrt{n - m}  is an integer.

Statement 2  is sufficient.

Answer: B

DeeptiManyaExpert · Answer

This is a Yes-No DS question. Start by using plugging in for such questions.

m,n --> Positive integers Is √(n−m) an integer?

1) n > m + 15
n - m > 15

n - m = 16 Answer to the question [Is √(n−m) an integer?] -- Yes
n - m = 17 Answer to the question [Is √(n−m) an integer?] -- No

Insufficient

2) n = m(m+1)

m = 1, n = 2 => √(n−m) = 1 Answer to the question [Is √(n−m) an integer?] -- Yes
m = 2, n = 6 => √(n−m) = 2 Answer to the question [Is √(n−m) an integer?] -- Yes
m = 3, n = 12 => √(n−m) = 3 Answer to the question [Is √(n−m) an integer?] -- Yes
m = 6, n = 42 => √(n−m) = 6 Answer to the question [Is √(n−m) an integer?] -- Yes

Sufficient

Thus, the answer to the question is B.

Hope that helped! Point to remember - Keep Math simple. Don't complicate it!

Plugging In can be a really helpful tool in some of the superficially complicated questions.

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If m and n are positive integers, is root (n-m) an integer?

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GMAT Data Sufficiency (DS) Questions

If m and n are positive integers, is root (n-m) an integer?

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