Is a positive integer x square of an integer?

Question

Is a positive integer x square of an integer?
A) x has exactly 2 different prime factors
B) x has exactly 12 positive divisors including 1 and x.

nik256 · Accepted Answer

This question is really interesting. we need to find if X is a perfect square.(yes/no)

Lets look at St. 1

X has exactly 2 different prime factors

So we can take the 2 smallest prime factors say 2 and 3, so x is 6, (no) 
 but if we take the other no like 36 it has 2 different prime factors and is a sq, (6)^2(yes)

We have a yes and a no to st1 so eliminate A and D

lets go to st 2

We know that X has 12 positive factors including one and x

We know the property of perfect squares, i.e they have odd factors.

So we know that x is not a perfect square,

(B) is the correct answer.

broall · Answer

A)  x=2 	imes 3=6  has extractly 2 prime factors but 6 is not square of an integer.
 x=2^2 	imes 3^2=36  has extractly 2 prime factors and 36 is square of an integer.
Insufficient.

B) Sufficient.

Let's assume that  x=p_1^{q_1} 	imes p_2^{q_2} 	imes ... 	imes p_n^{q_n} , where  p_1, p_2, ..., p_n  is distinctive prime factors.

The extractly number of factors (divisors) of  x  is  (q_1+1)(q_2+1)...(q_n+1)

From (B), we have  (q_1+1)(q_2+1)...(q_n+1)=12 . Let solve for each possible case:

If  12=12 , we have  q_1+1=12 \iff q_1=11 . So  x=p_1^{11} , this is not square of an integer (the power is odd). 
 If  12=2 	imes 6 , we have  q_1+1=2  and  q_2+1=6  (No need to care about the order). So  q_1=1  and  q_2=5 . Hence  x=p_1^1 	imes p_2^5 , this is not square of an integer. 
 If  12=3 	imes 4 , we have  q_1+1=3  and  q_2+1=4  (No need to care about the order). So  q_1=2  and  q_2=3 . Hence  x=p_1^2 	imes p_2^3 , this is not square of an integer. 
 If  12=2 	imes 2 	imes 3 , we have  q_1+1=2  and  q_2+1=2  and  q_2+1=3  (No need to care about the order). So  q_1=1  and  q_2=1  and  q_3=2 . Hence  x=p_1^1 	imes p_2^1 	imes p_3^2 , this is not square of an integer.

For all cases, we have  x  is not a square of an integer. So sufficient.

The answer is B

IanStewart · Answer

Statement 1 is not sufficient - as in the other reply, x can be 36 or 6, each of which have two prime divisors, so may or may not be a perfect square.

Perfect squares always have an odd number of divisors, so if Statement 2 is true, and x has an even number of divisors, x cannot be a perfect square, so Statement 2 is sufficient.

If it's not clear why perfect squares must have an odd number of divisors, you can think about listing out the divisors for different numbers. If you take, first, a number like 15, which is not a square, the divisors will come in pairs that produce 15 as a product:

1 and 15
3 and 5

so we have an even number of divisors. It is only for perfect squares that we have a divisor that is not in a pair (since we'd need to pair it with itself) - so for example, the divisors of 36 are

1 and 36
2 and 18
3 and 12
4 and 9
6

and since we have that single 'unpaired' divisor, 36 has an odd number of divisors in total. And the same will be true for any other square.

Is a positive integer x square of an integer?

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GMAT Data Sufficiency (DS) Questions

Is a positive integer x square of an integer?

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