Let m and n be positive integers. Is (m^13 - m^12)/n an integer?

Question

(1) m = n+1 
(2) m^2 = 5 - n^2

divyadna · Answer

Statement I is clearly sufficient.
Reject B,C, and E.

Statement II
m^2=5-n^2
m^2+n^2=5
Since it is stated in the question stem that m and n are positive integers, there are only two possibilities :-
1) m= 2, n=1
4+1=5
In this case (m^13-m^12)/n= (2^13-2^12)/1= 2^12(2-1)=2^12=integer.
Answer is YES.

2) m= 1, n=2
1+4=5
In this case (m^13-m^12)/n= 1^13-1^12/2= 0=integer.
Answer is YES again.

So, statement II alone is sufficient.
Ans is option D

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Crater · Answer

Whether  \frac{{m^{12}(m-1)}}{{n}}=Integer?

m=n+1

\frac{{(n+1)^{12}(n+1-1)}}{{n}}={(n+1)}^{12}

Hence, Sufficient

m^2+n^2=5

Possible Pairs of m and n (+ and - values both) are:
 1^2+2^2=5 
 2^2+1^2=5 
 0^2+5^2=5 
 5^2+0^2=5

For all the pairs we can say  \frac{{m^{12}(m-1)}}{{n}}=Integer

Hence sufficient, opt (D)

chandnik3 · Answer

In this, isn't the combination 5,0 or 0,5 wrong? as it has been said that m and n are positive numbers.
Therefore, 0 is not possible?
Also kindly clarify my understanding that m^13-m^12 will be divisible by 2 because if m is odd then m^13 and m^12 both will result in odd no. and Odd-Odd= Even.
And m is even then m^13 and m^12 both will result in even no. and even-even= Even.

Thanks in advance.

chetan2u · Answer

Let m and n be positive integers. Is  \frac{(m^{13 }- m^{12})}{n}  an integer?

\frac{(m^13 - m^12)}{n}=\frac{m^{12}(m-1)}{n} 
So n should be a factor of m or m-1

(1) m = n+1 
So n=m-1. This tells us that n is a factor of m-1.
Sufficient

(2)  m^2 = 5 - n^2 
 m^2+n^2=5 
Since both m and n are positive integers, only possibility is  1^2+2^2=5 
Two cases-
m=1 and n=2 : This will make m-1=0 a r the entire term 0.
m=2 and n=1 : This will make the denominator 1, and again the expression will be an integer. 
Sufficient

D

Let m and n be positive integers. Is (m^13 - m^12)/n an integer?

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GMAT Data Sufficiency (DS) Questions

Let m and n be positive integers. Is (m^13 - m^12)/n an integer?

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