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

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)

target
is \(\sqrt{k}\) an integer

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

possible value of k is 1 or 0
insufficient
#2
The tens digit of k is a factor of a single digit prime number
11 , no to target
16 yes to target
insufficient
from 1 &2 we can say that k is 1
sufficient
option C

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

(1) k is a multiple of every single-digit prime number.
k = 210,420,630.....210*210
if k = 210, sqrt k is not an Integer, if k = 210*210, sqrt k is an Integer.. not sufficient

(2) The tens digit of k is a factor of a single digit prime number.
no sufficient

1 + 2
1 is the only common number which can be factor of any of prime numbers
if 1 is in tens digit then the number cannot be a perfect square
hence, C

Asked: If k is a positive integer, is \(\sqrt{k}\) an integer ?

\(\sqrt{k}\) is an integer only if k^2 is a perfect square.

(1) k is a multiple of every single-digit prime number.
k = 2*3*5*7x = 210x ; where x is an integer
If k = 210^2 ; \(\sqrt{k} = 210\) is an integer
But if k=210; \(\sqrt{k} = \sqrt{210}\) is NOT an integer
NOT SUFFICIENT

(2) The tens digit of k is a factor of a single digit prime number.
Tens digit of k is either 1 or a single digit prime number
Tens digit of k is 1 or 2 or 3 or 5 or 7
If k=36; \(\sqrt{k} = 6\) is an integer
But if k=55; \(\sqrt{k} = \sqrt{55}\) is NOT an integer
NOT SUFFICIENT

(1) + (2)
(1) k is a multiple of every single-digit prime number.
k = 2*3*5*7x = 210x ; where x is an integer
(2) The tens digit of k is a factor of a single digit prime number.
For k to be an integer k is to be of the form = 210^2x^2 = 44100x^2
Last 2 digits of k is to be 00 for k to be an integer
Since tens digit is 1 or 2 or 3 or 5 or 7 and NOT 0.
k is NOT an integer
SUFFICIENT

IMO C

Imo C

Statement 1: k is a multiple of every single-digit prime number.
k Can be 2*3*5*7 or k can be (2*3*5*7)^2
Clearly Insufficient

Statement 2: The tens digit of k is a factor of a single digit prime number.
K can be 10 or 25
Insufficient

Combining both the only value k can take is 2*3*5*7

Using statement 1,
the smallest number possible is 210, and rt(210) is not an integer. But if the number was say 44100, which is a multiple of every single-digit prime number, rt(44100) = 210, which is a positive integer. So we cannot say for sure.

Using statement 2,
let k = 36. rt(36) is 6.
let k = 24. rt(24) is not an integer.
whereas both 2 and 3 are factors of a single digit prime number. Hence, statement 2 isn't enough either.

Using both the statements,
we can conclude that k is a number which is a multiple of 210. It won't have a root until k is 44100, whose tens digit is not a factor of a single digit prime number. Hence, rt(k) is not an integer.

Option C is the answer.

Correct answer : Choice C

statement 1 basicially says : k = 2*3*5*7*x

if x also equals 2*3*5*7 then ,yes k becomes a perfect square and hence root k will be an integer. If K is anything else, root k wont be an integer . Hence choice A and D are out

Statment 2 basically says tens digit of k is a prime number : lets assume k = 36, then yes root k will be an integer .
if k = 33, then root k wont be an integer and hence Option B is out

Now between choice C and E, using both statements we can confidently conclude that , k is not a perfect square and hence root k wont be an integer. Hence choice C is the right answer

1. Tricky wording. This st basically means that k is a multiple of LCM(2,3,5,7)=210 and hence it is just not equal to 210. Since this is the case, k could be 210 or 210^2 and hence this is NS
2. We do not know the number so we cannot determine

1 and 2:
For k to be a perfect square the last two digits must be 0 and since we know tens digit is a factor of a prime number we can be sure it is not a perfect sqaure. Suff
Ans C

1. k is a multiple of every single-digit prime number means K is a multiple of 2,3,5 & 7. Take LCM, it will be 210. K is a multiple of 210, i.e. 210, 420, 630..... etc. It is not sufficient as 210 * 210 will be perfect square & hence \sqrt{k} is Integer , but rest not.

2. The tens digit of k is a factor of a single digit prime number. Again not sufficient.

Combine both: integer with prime number at tens will not be a perfect square.

Hence C is the answer

(1) k is a multiple of every single-digit prime number.
Insufficient
Eg: (i) k = 2 x 3 x 5 x 7 -> Not a square
(ii) k = 2^2 x 3^2 x 5^2 x 7^2 -> Square

(2) The tens digit of k is a factor of a single digit prime number.
Insufficient
Eg: (i) k = 17 -> Not a square
(ii) k = 16 -> Square

Together,
We see that the tens digit must be 1, 2, 3, 5, 7 to be a factor of a single digit prime.
It cannot be zero, which implies that 2^2 x 5^2 is not present in the factorisation, thereby showing that the number cannot be a perfect
square.

=> Option C.

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

a) k = (2^a)*(3^b)*(5^c)*(7^d)

a,b,c,d are greater than 1. also if a,b,c,d are multiple of two then sqrt(k) will be a perfect integer, abcd all are equal to 1, then sqrt(k) will not be a perfect integer
hence a alone is not sufficient

b) possible values of tens digit of k are 1,2,3,4,5,6,7,8,9.

hence 2 alone is not sufficient

a and b

we know that for k to be perfect square it should have at least a, and b equal to 2 i.e. 2^2, and 5^2 which = 100,

also as per statement 2 tens digit cannot be 0, hence we conclude that k will not be a perfect square.

Answer should be C.

\sqrt{k} can be an integer only if k is a perfect square.

Single digit prime numbers are 2, 3, 5 and 7.
Statement 1: If k is a multiple of all 4 nos above, it is also a multiple of their LCM which is 210. So, k can be 210 (ans no) or can be square of 210 (ans yes). No single answer, hence insufficient statement.

Statement 2: Tens digit of k can be 1. So, k can be 15 (ans no) or 16 (ans yes). No single answer, hence insufficient statement.

Combining both statements, in order to be a perfect square, there should be even no. of zeros in the end. However, this is not possible due to constraint of statement 2. Hence, the answer will always be "no" and answer would be C.

(1) k is a multiple of every single-digit prime number.
then k is multiple of 2, 3, 5, and 7 then k may be 210 and Sqrt (210) is not an integer,
and k may also be 44100 and Sqrt (44100)=210 which is an integer,
So 1st statement is not sufficient,
(2) The tens digit of k is a factor of a single digit prime number.
the only number that is a factor of a single digit prime number is 1.
then the tens digit of K is 1,
We could have k =2116 =46^2,
but we could have k=210
So clearly not sufficient,

Let's take into account both statements, in this case we'll hold just one number which is 210, then, Sqrt (210) is not an integer,
So answer is C

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..

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.