Last visit was: 23 Apr 2026, 14:50 It is currently 23 Apr 2026, 14:50
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
MaggieSG
Joined: 01 Feb 2017
Last visit: 28 Apr 2017
Posts: 1
Own Kudos:
21
 [21]
Posts: 1
Kudos: 21
 [21]
3
Kudos
Add Kudos
18
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
singhabhijeet
Joined: 02 Aug 2013
Last visit: 05 Oct 2018
Posts: 49
Own Kudos:
32
 [6]
Given Kudos: 81
Location: India
Concentration: Technology, Leadership
WE:Programming (Consulting)
Posts: 49
Kudos: 32
 [6]
4
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
General Discussion
avatar
Andrenico
Joined: 20 Jan 2017
Last visit: 28 Mar 2017
Posts: 1
Own Kudos:
1
 [1]
Given Kudos: 1
Posts: 1
Kudos: 1
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
User avatar
grichagupta
Joined: 13 Dec 2016
Last visit: 02 Sep 2018
Posts: 32
Own Kudos:
35
 [1]
Given Kudos: 570
Posts: 32
Kudos: 35
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
MaggieSG, Thank you for posting this question. I made a mistake of considering x =6, y=0 and z=0 only to realize that the question clearly states POSITIVE INTEGERS and 0 is neither positive nor negative. Hence, lesson learnt. Need to keep this fact in mind while solving questions such as this.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
810,857
 [3]
Given Kudos: 105,853
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,857
 [3]
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
MaggieSG
p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are positive integers and x+y+z = 6, what is the smallest possible value of p?

(A) 64
(B) 240
(C) 360
(D) 640
(E) 900

Hi, I was working on this questions below and I assumed that x, y and z had to be 3 different numbers. But apparently I was wrong. I haven't been using math for a long time, but I remember we would always use different letters to refer to different values. So I thought: 1+2+3=6 was correct.
However, based on the final answer, it seems that the correct assumption is x+y+z = 6 = 1+1+4=6. So x= 1 and y also =1

I would be wrong then if I assume in other math problems that different letters refer to different values?
Thanks!

Unless it is explicitly stated otherwise, different variables CAN represent the same number.
avatar
sillyboy
Joined: 04 Dec 2016
Last visit: 19 Mar 2017
Posts: 37
Own Kudos:
128
 [1]
Given Kudos: 25
Location: India
GPA: 3.8
WE:Operations (Other)
Products:
Posts: 37
Kudos: 128
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
+1 B.

x=4
y=1
z=1

=> 2^x*3^y*5^z = 2^4 * 3^1 * 5^1 = 16*3*5 = 240
User avatar
devctg
User avatar
Wharton Moderator
Joined: 30 May 2015
Last visit: 10 Nov 2025
Posts: 30
Own Kudos:
39
 [2]
Given Kudos: 103
Products:
Posts: 30
Kudos: 39
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
x + y + Z = 6
we can chose bigger value for x , as it is power of 2 which is smallest number out of given three values

so x =4 ; y =1 ; z = 1

then 2^x 3^y 5^z => 16*3*5 = 240
User avatar
guptarahul
Joined: 27 Mar 2017
Last visit: 31 Jan 2018
Posts: 56
Own Kudos:
Given Kudos: 24
Status:Gmat lover
Location: India
GMAT 1: 710 Q49 V39
GPA: 3.91
Products:
GMAT 1: 710 Q49 V39
Posts: 56
Kudos: 25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
MaggieSG
p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are positive integers and x+y+z = 6, what is the smallest possible value of p?

(A) 64
(B) 240
(C) 360
(D) 640
(E) 900

Hi, I was working on this questions below and I assumed that x, y and z had to be 3 different numbers. But apparently I was wrong. I haven't been using math for a long time, but I remember we would always use different letters to refer to different values. So I thought: 1+2+3=6 was correct.
However, based on the final answer, it seems that the correct assumption is x+y+z = 6 = 1+1+4=6. So x= 1 and y also =1

I would be wrong then if I assume in other math problems that different letters refer to different values?
Thanks!

As we want to make number as small as possible, so we will take power of smallest number i.e 2, as greater as we can take.
so we take,
x=4 :)
y=1 :)
z=1 :)

x+y+z=6

=> 2^x*3^y*5^z = 2^4 * 3^1 * 5^1 = 16*3*5 = 240
User avatar
rever08
Joined: 21 Jul 2017
Last visit: 13 Jan 2020
Posts: 148
Own Kudos:
Given Kudos: 143
Location: India
Concentration: Social Entrepreneurship, Leadership
GMAT 1: 660 Q47 V34
GPA: 4
WE:Project Management (Education)
Products:
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Remember words play important role. since x, y, and z are positive they cannot be 0.
User avatar
sashiim20
Joined: 04 Dec 2015
Last visit: 05 Jun 2024
Posts: 608
Own Kudos:
Given Kudos: 276
Location: India
Concentration: Technology, Strategy
WE:Information Technology (Consulting)
Kudos
Add Kudos
Bookmarks
Bookmark this Post
MaggieSG
p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are positive integers and x+y+z = 6, what is the smallest possible value of p?

(A) 64
(B) 240
(C) 360
(D) 640
(E) 900
\(p = (2^x )( 3^y)(5^z)\)

\(x + y + z = 6\)

\(x, y,\) and \(z\) are positive integers.

Smallest value of '\(p\)' would be when \(x, y , z\) would be smallest positive integers.

\(x = 4\)
\(y = 1\)
\(z = 1\)

\(x+y+z = 4 + 1 + 1 = 6\)

\(p = (2^4)( 3^1)(5^1) = (16)(3)(5) = 240\)

Answer (B)...
User avatar
vasu1104
Joined: 10 Feb 2023
Last visit: 23 Apr 2026
Posts: 391
Own Kudos:
Given Kudos: 664
Location: Canada
Products:
Posts: 391
Kudos: 236
Kudos
Add Kudos
Bookmarks
Bookmark this Post
i found the easiest way is to factorize the given choices and see which has 2,3 and 5 as a factor and which one is smallest in all.
by doing this 240 is smallest
MaggieSG
p is equal to the product of 2^x, 3^y, and 5^z. If x, y, and z are positive integers and x+y+z = 6, what is the smallest possible value of p?

(A) 64
(B) 240
(C) 360
(D) 640
(E) 900

Hi, I was working on this questions below and I assumed that x, y and z had to be 3 different numbers. But apparently I was wrong. I haven't been using math for a long time, but I remember we would always use different letters to refer to different values. So I thought: 1+2+3=6 was correct.
However, based on the final answer, it seems that the correct assumption is x+y+z = 6 = 1+1+4=6. So x= 1 and y also =1

I would be wrong then if I assume in other math problems that different letters refer to different values?
Thanks!
User avatar
arushi118
Joined: 21 Jul 2024
Last visit: 19 Apr 2026
Posts: 267
Own Kudos:
Given Kudos: 894
Location: India
Concentration: Leadership, General Management
GPA: 8.2/10
Products:
Posts: 267
Kudos: 76
Kudos
Add Kudos
Bookmarks
Bookmark this Post
p = 2^x.3^y.5^z
x+y+z = 6
z = 1 (as x, y, z are positive - none of them can be 0)

x,y -> 1,4 or 4,1 or 2,3 or 3,2
if we try with x = 4, y=1 and z=1 : we get p = 240 which is the smallest number possible.
Moderators:
Math Expert
109785 posts
Tuck School Moderator
853 posts