HOT Competition 27 Aug/8PM: If k is a positive integer, is k^(1/2) an

Question

If k is a positive integer, is  \sqrt{k}  an integer?

(1) The number of positive factors of k is a prime number
(2) k has a factor p such that 1 < p < k

M36-102

Bunuel · Accepted Answer

Official Explanation

If k is a positive integer, is  \sqrt{k}  an integer?

Notice that the question asks whether k is a perfect square (the square of an integer). Recall that the number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square.

(1) The number of positive factors of k is a prime number.

If the number of factors of k is 2, then that would mean that k is a prime, so  \sqrt{k}  will NOT be an integer.
If the number of factors of k is any other prime, then that would mean that the number of factors of k is odd, and therefore k is perfect square, so  \sqrt{k}  will be an integer.

Not sufficient.

(2) k has a factor p such that 1 < p < k. This is another way of saying that k is NOT a prime number (because if k were a prime, it would not have any other factor but 1 and itself). Still not sufficient, consider k = 4 and k = 6.

(1)+(2) From (2) we have that k is NOT a prime number, so we are left with second case from (1): the number of factors of k is odd, and therefore k is perfect square, so  \sqrt{k}  is an integer. Sufficient.

Answer: C.

sk7075 · Answer

k = +ve I
Q. sqrt k = Int
St.1
k = 2 
Factors = 1 and 2 (Prime) but sqrt not I. No
k=4 Yes
Insufficient

St.2 
k is not equal to prime
k= 15 No
k=16 Yes
Insufficient

Combining I and II
We get a definite answer that sqrt k = I 
Hence C

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minustark · Answer

If k is a positive integer, is an integer?

(1) The number of positive factors of k is a prime number
(2) k has a factor p such that 1 < p < k

1) if k has just 2 factors, then square root k is not an integer. for example: when k = 2 or 3.
But when k has 3 or 5 factors, then square root k is an integer. Example square root of 4 or 16. Not sufficient.

2) k is not a prime integer. Clearly not sufficient. Square root of 6 not an integer but square root of 9 is an integer. Not sufficient

Together, the number of factors of a square of an integer is always odd. And as we can infer the number not factors of k cannot be 2, so in every case square root k will be an integer. Sufficient

C is our answer.

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Anki2609 · Answer

K is an integer  \sqrt{k}  is an integer or not. 
We have to find if k is square of any number. 
All the number has factor pairs except squares. They always have odd number of factors.

Statement 1:
Number of factors are prime number. 
Except 2 all the prime numbers are odd. 
So we have 2 options either k is square or k is prime(in that case it will have 2 factors) :  insufficient

Statement 2:
p is a factor of k and it is between 1 and p so we know that k is not prime. But still we don't know if k is square:  insuff

Combining 1+2
We know that k has odd number of factors. 
So, C is sufficient.

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Saupayan · Answer

In other words, the question is asking is k the square of an integer?

This means k has either 2 factors or an odd number of factors.
Only prime numbers have 2 factors. In that case, k isn't the square of a number and √k isn't an integer.
If k has odd number of factors, it means k is the even power of a number. Which means, k is also the square of a number and √k is an integer.
Hence, (1) is not sufficient

This means that k has more than 2 factors. So, it has 3 or more factors. (Now, I don't know how literally we should take "a factor". Does it mean "exactly one" or "at least one"? If "exactly one", then it means k has exactly 3 factors, meaning k is the square of a prime number, and statement 2 will be sufficient. But, I chose to go with the 2nd interpretation, i.e. "k has at least 1 factor between 1 and k")
6 has at least 1 factor between 1 and 6, and it's not a square
9 has at least 1 factor between 1 and 9, and it is a square.
So, clearly statement (2) isn't sufficient either.

Taking (1) and (2) together:
We know k doesn't have 2 factors. It has more than 2 factors. So, it can have only odd number of factors.
As discussed under (1), if it has odd number of factors, it means that k is the even power of an integer. That means, k is a square and √k is an integer.

(1) and (2) are together sufficient to answer the question.
Answer: C

piyush0122 · Answer

If k is a positive integer, is √k an integer?

(1) The number of positive factors of k is a prime number- insufficient as k can 2, factors of 2 are 1,2. so (k)^1/2 is not an integer. k can be 4. factors of 4 are 1,2, & 4. so (k)^1/2 is an integer.
(2) k has a factor p such that 1 < p < k. insufficient as k can be 4 and the given condition will be true or k can be 8 and the given condition will not be true.

Together taking (1) & (2) k can be 4, 9, 16, 25 etc. so (k)^1/2 will always be an integer.
so c is sufficient.

GMATinsight · Answer

ANswer: Option C

https://www.youtube.com/watch?v=Rx32Wdu-TJw

bumpbot · Answer

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HOT Competition 27 Aug/8PM: If k is a positive integer, is k^(1/2) an

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HOT Competition 27 Aug/8PM: If k is a positive integer, is k^(1/2) an

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