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

Question: Is \(\sqrt{k}\) an integer ?

S1:
"k is a multiple of every single-digit prime number."
So k=2*3*5*7*n = 210*n [n is an integer > 0]
This means k can be equal to 210, in which case \(\sqrt{210}\) is NOT an integer
and k can also be equal to 210*210=44100 in which case \(\sqrt{44100}\) is an integer
So, INSUFFICIENT

S2:
"The tens digit of k is a factor of a single digit prime number."
So possible tens digit of k could be 1,2,3,5,7
Consider k=51 \(\sqrt{51}\) is NOT an integer
again consider k=25, here \(\sqrt{25}\) is an integer
So, INSUFFICIENT

S1 + S2:
For \(\sqrt{k}\) to be an integer, k has to be a square number. In order to satisfy that k is a square number and a multiple of 2,3,5 and 7 the minimum possible value is 210*210=44100. Here the tens digit is 0 which is not a factor of single digit prime. To get the next square number (k) we have to multiple 44100 with 2^2*3^2*5^2*7^2. But clearly this will lead to a 0 in tens digit, and so will all other subsequent squares. But this is against statement 2. So K cannot be a square number and \(\sqrt{k}\) is NOT an integer.
So together SUFFICIENT

Ans C

If k is a positive integer, is k√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.

from Question stem,

k is always an integer (Given)..we have to find out about root(k)

!1) k is a multiple of every single-digit prime number.
k = 2*3*5*7 a (where 'a' is a constant)
= 210 a

However, root (k) can not be integer when a = 1
root (k) can be an ineger when a =210

Since, both YES & NO, unique answer NOT possible

so, (A) and (D) are out

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

YES for k =121, since root (k) = 11 (an integer)
NO for k = 120, since root (k) = NOT an ineger

Since, both YES & NO, unique answer NOT possible
so, (B) is out

(1) & (2)

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

k will be of the form 210a such that k= 420, 630, 1050, 1470 etc

for all such values, root (k) is NOT an integer
Hence, NO is the unique answer for his case

thus, E is out

We get

(C) is the CORRECT answer

Hope this helps..

Answer: A

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

(1) k is a multiple of every single-digit prime number.
Single-digit prime numbers are 2, 3, 5, 7
=> k = 2*3*5*7
=> k^(1/2) is not an integer.
Sufficient.

(2) The tens digit of k is a factor of a single digit prime number.
Let k = 36
Here, tens digit 3 is a factor of a single digit prime number.
k^(1/2) = 6 --> Integer
Let k =35
then k^(1/2) is not an integer.
Not sufficient.

I chose C.

I suspect the answer will be C or E. I usually choose C when I am confused with C or E.

Statement 1. Option A is not the answer K will be a positive multiple of 210 or a positive multiple of (2*3*5*7).
Therefore K=210M where M is a positive integer. Not Sufficient.

Statement 2. Option B is not the answer. Assume K is 25 or K is 28. Therefor Not Sufficient.

Therefore Option D cannot be the answer.

The answer will be option C or E. I guessed C.

If k is a positive integer, is √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.

1. We have k is a multiple of 2,3,5,7. But we are not told how many factors are there. It could be 2^2*3^2*5^2*7^2 => insuf
2. Ten digits can be 1,2,3,5,7, it can be any number => insuf

With both, we now know the √k is integer when at least we have 2^2*3^2*5^2*7^2 => this result will have the last two number 00 (cause we have 2^2*5^2) => This will never happen because we need the ten digits 1,2,3,5, or 7
=> So the number will not be integer, suff. C

My answer is E for this question

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

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

From Statement one we know that k = 2*3*5*7*a it means the value of K depends upon value of a
Now if a=5 K is not a perfect square, but if a=210 then K is a perfect square . Two answers so insufficient

(2) The tens digit of k is a factor of a single digit prime number.
if k = 100 tens digit is a factor of single digit prime number here K is a perfect square but when k = 300 it's not a perfect square
Indufficent

Combinig both
If k = (210) raise to 2 then tens digit will be 0 that is factor of single digit prime number and hence k is a percfect square
But if K = 210 * 10 then also tens digit will be 0 that is factor of single digit prime number but k is not a percfect square

Answer is E

If k is a positive integer, is √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.

IMO ans. A

1) k=1
2) kk=1k unit digit of 1k falls from 1 to 9 which is insufficient to estimate

k has to be a square of an integer

(1) k is a multiple of every single-digit prime number. Smallest such square is 44100
k can be 4x9x25x49=44100 or 2x3x5x7=210.
Insufficient
(2) The tens digit of k is a factor of a single digit prime number.
Tens digit can be any of 1, 2, 3, 5, 7, since factor of a prime number are one and itself. This doesnot tell us anything about k being square or not. For eg last two digits can be 10 or 2.. can be 210 or 420
Insufficient

Both
Smallest such square will be 44100. Any multiple of this number will have to have 0 at tens place, be it square or not. But by 2, tens digit is a factor of single digit prime number, which is not possible. So k is not a square. Sufficient

Ans C

Given information: k: Positive integer

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

k is multiple of {2,3,5,7}, which implies that k is a multiple of LCM of all those numbers => k is a multiple of 210

If k=210, then root(k) NOT AN INTEGER
If k=210*210 then root(k) YES AN INTEGER

NOT SUFFICIENT

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

Tens Digit (TD) = 1/2/3/5/7

If k=25, then root(k) YES AN INTEGER
If k=35, then root(k) NOT AN INTEGER

NOT SUFFICIENT

(1)+(2) Together

k is a multiple of 210
k has a tens digit of 1/2/3/5/7

Now for k to be a multiple of 210 and for root(k) to be an integer, k will have to equal 210*210 (because 210=2*3*5*7 and only way and least possible number for perfect square is 2^2 * 3^2 * 5^2 * 7^2 which is 210^2).
But 210^2 will have a tens digit of 0 and that does not match with our statements combined together
So it is not possible
Root(K) is not an integer

SUFFICIENT

Answer - C
Answer - C

The single digit prime numbers are 2, 3, 5, and 7.

Statement 1: K is a multiple of every single digit prime number.

This means that K is a multiple of each of 2, 3, 5, and 7. This means K must a multiple of the LCM of 2, 3, 5, and 7 - which is 210.

Therefore, K is of the form = 210p where p is any positive integer.

Case 1: p = 1 therefore K = 210 which is not a perfect square. Hence, √K is not an integer.

Case 2: p = 210 therefore K = 210*210 = 44100 which is a perfect square. Hence, √K is an integer.

Statement 1, therefore, is insufficient. Eliminate A and D.

Statement 2: The tens digit of K is a factor of a single digit prime number.

The single digit prime numbers are 2, 3, 5, and 7. And all possible factors of these are 1, 2, 3, 5, and 7. The tens digit of K can be any of these five integers.

For ex: Let's say the tens digit of K is 2.

Case 1: K = 25, clearly K is a perfect square and √K is an integer.

Case 2: K = 27, clearly K is not a perfect square and √K is not an integer.

Statement 2, therefore, is insufficient. Eliminate B.

Combining the two statements

K = 210p and tens digit of K is either 1 or 2 or 3 or 5 or 7.

In order for K to be a perfect square, each of its existing factors must first be squared. Which means p must at least be 210, and then can optionally be multiplied by Z where Z is a non-zero perfect square.

Which means for K to be a perfect square, K = 210*210*Z = 44100*Z where Z is a non-zero perfect square. Multiplying 44100 by any positive integer Z will result in two 0s at the end.

Which means for K to be a perfect square, the tens digit of K ought to be 0 and nothing other than that. However, we've been given from statement 2 that the tens digit of K is either 1 or 2 or 3 or 5 or 7.

Hence, clearly, upon combining the two statements we can say that K is not a perfect square and √K is not an integer. Hence, option C.

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IMO-C



(1) k is a multiple of every single-digit prime number.
if k=. 2*3*5*7 - root k not a integer
if :- k=. (2*3*5*7)^2 - root k is a integer
Not sufficient


(2) The tens digit of k is a factor of a single digit prime number.
It says ten digits is equal to 1

Both C
K = 210
Sufficent

Statement (1)
Let's say k = 210 then √k is not an integer
But if k = 44100 then √k is an integer

Insufficient

Statement (2)
Let's consider 2 as the tens digit as 2 is a factor of 2, a single digit prime number
For e.g. k = 25, then √k is an integer
But For e.g. k = 24, then √k is not an integer

Insufficient

Combining Statement (1) and (2) we get that the number should be a multiple of 210 and its tens digit should be 1, 2, 3, 5 or 7
Now for k to be a perfect square each of the prime numbers should be repeated at least once or in even multiples thereafter. Hence the last two digits would necessarily have to end in 00 as we have two sets of 5 and 2. In that scenario, the number's Tens digit cannot be 1, 2, 3, 5 or 7.

So combining we have an answer as NO

IMHO Option C

My answer is A.

(1) 2*3*5*7= 210
No.
Sufficient
(2) xyz
y is a factor of a single digit prime number. We can not know the number by its tens digits

The answer is A

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

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

Single digit prime numbers : 2, 3, 5 and 7

Therefore, K can be : 2x2x3x3x5x5x7x7 and √k is an integer

But k can also be 2x3x5x7 and √k is NOT an integer

Statement 1 is INSUFFICIENT

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

K can be : 121 and √k is an integer
K can also be 122 and √k is NOT an integer

Statement 2 is INSUFFICIENT

Combining statements 1 and 2
k is a multiple of every single-digit prime number and The tens digit of k is a factor of a single digit prime number.


Here Both statements can not hold true. For k to be a multiple of every single-digit prime number as well as a perfect square, it must ende with two zeros i e (00) which will make our second condition invalid.

Ans: Option C

[quote="Bunuel"]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.

[quote="Bunuel"]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.


1) - Single digit prime numbers are - 2,3,5,7
Now lets consider k=2*3*5*7 =2100,Then in that case \(\sqrt{k}\) is not an integer.
Another example if k=2^2*3^2*5^2*7^2=44100 then in that case \(\sqrt{k}\) is an integer.Thus 1 is insufficient.
A and D out
2)Factor of a single digit prime number can be 1 nd the number itself(2,3,5,7).
Now for case like k=25,36,16 \(\sqrt{k}\) is an integer
And for case like 15,72,39 \(\sqrt{k}\) is not an integer.
Thus 2 is insufficient.B is out.
Combining 1 and 2
Now for the 2nd condition to be true the ten's digit has to be either of these digits (1,2,3,5,7)
But earlier we have seen for k to be integer the minimum value it has to be =2^2*3^2*5^2*7^2 which is 44100.
and all its multiple would have 0 at the tens place.
Thus \(\sqrt{k}\) cannot be an integer.
Thus the correct choice is C

(1) is not sufficient
(2) is not sufficient
Therefore it's answer E.

Explanation
(1) k = 2*3*5*7*n=210n
If n = 210, then square root of n is an integer, if n=3 for eg, it's not.
Not sufficient.

(2) The tens digit of k is a factor of a single digit prime number.
If so, then the tens digit of k is either 2 or 3 or 5 of 7.
That information isn't sufficient to determine if Sr of k is an integer.

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