GMAT Club World Cup 2022 (DAY 4): If k is a positive integer, is

Question

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

(1) k is a multiple of every single-digit prime number.
(2) The tens digit of k is a factor of a single digit prime number.

M38-10

Bunuel · Accepted Answer

GMAT CLUB Official Explanation:

If  k  is a positive integer, is  \sqrt{k}  an integer ? 
 
(1)  k  is a multiple of every single-digit prime number.
  
There are four single-digit prime numbers: 2, 3, 5, and 7. So,  k  is a multiple of  2*3*5*7 . This one is clearly insufficient, for example, if  k=2*3*5*7 , then the answer is NO but if  k=(2*3*5*7)^2 , then the answer is YES. Not sufficient.
 
Notice that, from this statement we can get that the units digit of  k  is 0.
   
(2) The tens digit of  k  is a factor of a single digit prime number.
  
There are four single-digit prime numbers: 2, 3, 5, and 7. So, the tens digit of  k  is 1, 2, 3, 5, or 7. This one is also clearly insufficient, for example, if  k=20 , then the answer is NO but if  k=25 , then the answer is YES. Not sufficient.
  
(1)+(2) We know from (1) that the units digit of  k  is 0. For  \sqrt{k}  to be an integer, the tens digit of  k  must also be 0 (so  k  must be divisible not only by 10 but also by  10^2=100 ). But from (2) we know that the tens digit of  k  is NOT 0, thus  \sqrt{k}  is NOT and integer. Sufficient.

Answer: C

KartikSingh09 · Answer

The questions asks whether k is a perfect square or not.

St 1: k can be translated as the LCM and multiples of LCM of (2,3,5 & 7) i.e. 210, 420, 630 and so on. 
 Here, k will only be an integer for all even powers of 2,3,5 & 7. Since we don't know exact value.   Insufficient

St 2: Since the factors of a prime number are 1 and the number itself k can be 16, 25, 26... as we don't know the exact
 value of k.   Insufficient

St 1 & 2 together : Statement 1 tells us that k is perfect square only when the powers of 2,5,3,7 are even, hence the minimum powers of 2&5 is '2' and the number will have 2 trailing zeroes( or 4,6,8..). Now, St 2 tells us that the tens digit has to be either 1 or any single digit prime number, therefore it cannot be '0', hence even powers of 2&5 are not possible. All the remaining numbers with odd powers will not be perfect squares.   Sufficient

Answer : C

Papai21 · Answer

From st. 1 we have, k is a multiple of every single-digit prime number
Thus k = 2 * 3 * 5 * 7 * constant (c) = 210 * c
If √k is an integer then we need the constant term c in k to be 
c = (2 * 3 * 5 * 7)^(odd powers)
But we are not given any information about c. So this statement is not sufficient.

From st. 2 we have, the tens digit of k is a factor of a single digit prime number
So tens digit of k could be = 1, 2, 3, 5, 7
If k = 25 or 36 then answer is yes. But if not then No.
So So this statement is not sufficient

Combining St 1 and St 2 we have
If √k is an integer then k must at least be 210 * 210 = 44100. Thus we see that units digit and tens digit of k must have 0 in order to √k to be an integer. But st 2 clearly says that the tens digit is not 0. Hence √k is not an integer.

So answer choice (C)

gmatophobia · Answer

Given

k is positive integer

Question

Is  \sqrt{k}  is an integer

Statement 1

k is a multiple of every single-digit prime number.

We know that k consists of 2 * 3 * 5 * 7

Case 1

If k = 2 * 3 * 5 * 7

Is  \sqrt{k}  is an integer - No

Case 2

If k =  2^2 * 3^2 * 5^2 * 7^2

Is  \sqrt{k}  is an integer - Yes

Therefore statement 1 is not sufficient

Statement 2

The tens digit of k is a factor of a single digit prime number.

So we know that the tens place of k is either 1 or 2 or 3 or 5 or 7

Case 1

k = 121

Is  \sqrt{k}  is an integer - Yes

Case 1

k = 120

Is  \sqrt{k}  is an integer - No

Therefore statement 2 is not sufficient

Combining

We know from statement 1 that k is a multiple of 210, now for  \sqrt{k}  to be an integer, k should be a multiple of  (210)^2  and if that's the case, k will violate condition 2.

Hence we can be sure that

\sqrt{k}  is an not integer 
 
 IMO C

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