Last visit was: 18 Nov 2025, 19:47 It is currently 18 Nov 2025, 19:47
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: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,086
 [8]
Given Kudos: 99,964
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
 [8]
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
Given Kudos: 99,964
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
Kudos
Add Kudos
Bookmarks
Bookmark this Post
General Discussion
avatar
Kanika3agg
Joined: 09 May 2018
Last visit: 20 Feb 2021
Posts: 96
Own Kudos:
Given Kudos: 75
Posts: 96
Kudos: 81
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
shridhar786
Joined: 31 May 2018
Last visit: 08 Feb 2022
Posts: 324
Own Kudos:
Given Kudos: 132
Location: United States
Concentration: Finance, Marketing
Posts: 324
Kudos: 1,718
Kudos
Add Kudos
Bookmarks
Bookmark this Post
power of 2 in 7! -- \(\frac{7}{2}\) - \(\frac{3}{2}\) - 1 -- 3+1 = 4 -- in \((7!)^2\) = 4*2 = 8

power of 3 in 7! -- \(\frac{7}{3}\) - 2 -- 2 -- in \((7!)^2\) = 2*2 = 4

power of 5 in 7! -- \(\frac{7}{5}\) - 1 -- 1 -- in \((7!)^2\) = 1*2 = 2

x = 8, y = 4, and z = 2

x+y+z = 8+4+2 = 14

D is the answer
avatar
nightvision
Joined: 04 Mar 2018
Last visit: 11 Sep 2020
Posts: 20
Own Kudos:
10
 [1]
Given Kudos: 34
GPA: 3.5
Products:
Posts: 20
Kudos: 10
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
7!^2 = (1*2*3*4*5*6*7)^2 = (2^4 * 3^2 *5^1*7^1)^2

so x= 8
y= 4
z = 2
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 18 Nov 2025
Posts: 21,712
Own Kudos:
Given Kudos: 300
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,712
Kudos: 26,990
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If x, y and z are largest positive integers for which \(2^x3^y5^z\) is a factor of \((7!)^2\), what is the value of \(x + y + z\) ?

A. 7
B. 12
C. 13
D. 14
E. 16

Since 7! = 7 x 6 x 5 x 4 x 3 x 2 = 7 x 2 x 3 x 5 x 2^2 x 3 x 2 = 2^4 x 3^2 x 5 x 7, then we see that (7!)^2 = 2^8 x 3^4 x 5^2 x 7^2. Therefore, x = 8, y = 4, z = 2 and hence x + y + z = 14.

Answer: D
User avatar
rollinbusiness
Joined: 02 Dec 2022
Last visit: 13 Nov 2023
Posts: 10
Own Kudos:
Given Kudos: 30
Posts: 10
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Solving it by writing out 7! makes sense to me, but I'm wondering why I got the wrong answer when I used the formula for prime factors of a factorial.


(7/2)+(7/4)+(7/8) (throw out remainder) = 3+1+0 = 4
(7/3)+(7/9) (throw out remainder) = 2+0 = 2
(7/5) + (7/25) (throw out remainder) = 1 + 0 = 1

4+2+1 = 7

So I got 7, but the answer should be 14? I'm confused.


Edit: I realized my error. The question says (7!)^2... I reread it so many times and didn't even notice the ^2... ugh...
Moderators:
Math Expert
105355 posts
Tuck School Moderator
805 posts