Last visit was: 22 Jul 2024, 22:32 It is currently 22 Jul 2024, 22:32
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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94517
Own Kudos [?]: 643138 [5]
Given Kudos: 86719
Send PM
Most Helpful Reply
VP
VP
Joined: 27 Feb 2017
Posts: 1472
Own Kudos [?]: 2326 [5]
Given Kudos: 114
Location: United States (WA)
GMAT 1: 760 Q50 V42
GMAT 2: 760 Q50 V42
GRE 1: Q169 V168

GRE 2: Q170 V170
Send PM
General Discussion
Intern
Intern
Joined: 05 Apr 2021
Posts: 10
Own Kudos [?]: 1 [1]
Given Kudos: 2
Send PM
Intern
Intern
Joined: 10 Jun 2023
Posts: 16
Own Kudos [?]: 12 [0]
Given Kudos: 8
Location: India
GMAT 1: 640 Q49 V29
WE:Engineering (Aerospace and Defense)
Send PM
Re: What is the number of positive integers that divide k^2 but do not div [#permalink]
zhanbo wrote:
A positive integer M that divides k^2 but does not divide k must satisfy
M = k * f where f is a positive factor (other than 1)of k.

(1) Only
A factor of 13 can only be obtained if k^2 = P^12 where P is a prime number.
So, k = P^6
The following six numbers can divide K^2 but not k: P^7, P^8, P^9, P^10, P^11, P^12.
Sufficient.

(2) Only
For √k to have a total of 4 factors, it is possible that √k = P^3 where P is a prime number, which is the same case as in (1). There should be six number.

It is, however, also possible to have a total of 4 factors if √k = P * Q where P and Q are two different prime numbers.
In this case, k=P^2*Q^2, k^2=P^4*Q^4.
The following eight numbers can divide K^2 but not k: P^2*Q^3, P^2*Q^4, P^3*Q^2, P^3*Q^3, P^3*Q^4, P^4*Q^2, P^4*Q^3, P^4*Q4.
(The eight can be calculated via 3*3-1 because we can select one from P^2, P^3, P^4 and another one from Q^2, Q^3, Q^4. Then we remove P^2*Q^2)

Not sufficient.

The answer is (A).

­Hi,
The solution posted above is almost correct but  I think you missed some other nos. which can divide P^4 * Q^4 for eg. P^3, P^4, Q^3 and Q^4 these nos. shall divide y^2 but not y. Please let me know if you agree.
GMAT Club Bot
Re: What is the number of positive integers that divide k^2 but do not div [#permalink]
Moderator:
Math Expert
94517 posts