Last visit was: 14 Dec 2024, 02:33 It is currently 14 Dec 2024, 02:33
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Dec 2024
Posts: 97,874
Own Kudos:
685,686
 []
Given Kudos: 88,270
Products:
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 97,874
Kudos: 685,686
 []
1
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Dec 2024
Posts: 97,874
Own Kudos:
Given Kudos: 88,270
Products:
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 97,874
Kudos: 685,686
Kudos
Add Kudos
Bookmarks
Bookmark this Post
General Discussion
User avatar
KartikSingh09
Joined: 14 Jun 2021
Last visit: 14 Dec 2022
Posts: 18
Own Kudos:
15
 []
Given Kudos: 17
Posts: 18
Kudos: 15
 []
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Papai21
Joined: 14 Aug 2021
Last visit: 07 Dec 2024
Posts: 40
Own Kudos:
59
 []
Given Kudos: 52
Location: India
Posts: 40
Kudos: 59
 []
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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)
User avatar
gmatophobia
User avatar
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 13 Dec 2024
Posts: 3,122
Own Kudos:
6,965
 []
Given Kudos: 1,860
Location: India
Concentration: Strategy, Leadership
Products:
Posts: 3,122
Kudos: 6,965
 []
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
Moderator:
Math Expert
97874 posts