BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Multiples

Expert replies
by joyseychow » Thu Aug 13, 2009 8:06 pm
If n is a multiple of 5, n=p^2q, where p and q are prime numbers. Which of the following must be a multiple of 25?

A. p^2
B. q^2
C. pq
D. p^2q^2
E. p^3q


Easy one, but I didn't get right. Why B cannot be the answer?
Join the discussion
Source: — Problem Solving |

by tohellandback » Thu Aug 13, 2009 8:45 pm
answer is D
n=p^2q
either p or q can be 5
clearly P^2*q^2 must be divisible by 25.

B cannot be the answer because
if p is 5 then q can be any prime number. 3,7, etc.
in that case p^2 will not be divisible by 25
The powers of two are bloody impolite!!
Join the discussion

by ankitns » Fri Aug 14, 2009 5:33 am
tohellandback wrote:answer is D
n=p^2q
either p or q can be 5
clearly P^2*q^2 must be divisible by 25.

B cannot be the answer because
if p is 5 then q can be any prime number. 3,7, etc.
in that case p^2 will not be divisible by 25

q cannot be 5...plug in the numbers...if p = 2 and q = 5...we get
n = 2^10 = 1024..not a multiple of 5....

n can only be a multiple of 5 only if p is 5...and if p is a multiple of 5 then p^2 must be a multiple of 25...

Hence the answer is A[spoiler][/spoiler]
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
Join the discussion

by tohellandback » Fri Aug 14, 2009 5:40 am
ankitns wrote:
tohellandback wrote:answer is D
n=p^2q
either p or q can be 5
clearly P^2*q^2 must be divisible by 25.

B cannot be the answer because
if p is 5 then q can be any prime number. 3,7, etc.
in that case p^2 will not be divisible by 25

q cannot be 5...plug in the numbers...if p = 2 and q = 5...we get
n = 2^10 = 1024..not a multiple of 5....

n can only be a multiple of 5 only if p is 5...and if p is a multiple of 5 then p^2 must be a multiple of 25...

Hence the answer is A[spoiler][/spoiler]
with p=2, how r u getting n=2^10
it will be n=2^2*5
..or hey may be you think its n=P^(2q)
The powers of two are bloody impolite!!
Join the discussion

by ankitns » Fri Aug 14, 2009 7:22 am
yes..I was thinking n=p^(2q)...

joyseychow: Could you please clarify whether the question states

n=p^(2q)

or

n=(p^2) * q


Thanks.

tohellandback wrote:
ankitns wrote:
tohellandback wrote:answer is D
n=p^2q
either p or q can be 5
clearly P^2*q^2 must be divisible by 25.

B cannot be the answer because
if p is 5 then q can be any prime number. 3,7, etc.
in that case p^2 will not be divisible by 25

q cannot be 5...plug in the numbers...if p = 2 and q = 5...we get
n = 2^10 = 1024..not a multiple of 5....

n can only be a multiple of 5 only if p is 5...and if p is a multiple of 5 then p^2 must be a multiple of 25...

Hence the answer is A[spoiler][/spoiler]
with p=2, how r u getting n=2^10
it will be n=2^2*5
..or hey may be you think its n=P^(2q)
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
Join the discussion

by joyseychow » Sun Aug 23, 2009 2:00 am
ankitns wrote:yes..I was thinking n=p^(2q)...

joyseychow: Could you please clarify whether the question states

n=p^(2q)

or

n=(p^2) * q


Thanks.

tohellandback wrote:
ankitns wrote:
tohellandback wrote:answer is D
n=p^2q
either p or q can be 5
clearly P^2*q^2 must be divisible by 25.

B cannot be the answer because
if p is 5 then q can be any prime number. 3,7, etc.
in that case p^2 will not be divisible by 25

q cannot be 5...plug in the numbers...if p = 2 and q = 5...we get
n = 2^10 = 1024..not a multiple of 5....

n can only be a multiple of 5 only if p is 5...and if p is a multiple of 5 then p^2 must be a multiple of 25...

Hence the answer is A[spoiler][/spoiler]
with p=2, how r u getting n=2^10
it will be n=2^2*5
..or hey may be you think its n=P^(2q)
Sorry. It's n=(p^2) * q.
Join the discussion

by real2008 » Mon Aug 24, 2009 11:05 am
joyseychow wrote:
ankitns wrote:yes..I was thinking n=p^(2q)...

joyseychow: Could you please clarify whether the question states

n=p^(2q)

or

n=(p^2) * q


Thanks.

tohellandback wrote:
ankitns wrote:
tohellandback wrote:answer is D
n=p^2q
either p or q can be 5
clearly P^2*q^2 must be divisible by 25.

B cannot be the answer because
if p is 5 then q can be any prime number. 3,7, etc.
in that case p^2 will not be divisible by 25

q cannot be 5...plug in the numbers...if p = 2 and q = 5...we get
n = 2^10 = 1024..not a multiple of 5....

n can only be a multiple of 5 only if p is 5...and if p is a multiple of 5 then p^2 must be a multiple of 25...

Hence the answer is A[spoiler][/spoiler]
with p=2, how r u getting n=2^10
it will be n=2^2*5
..or hey may be you think its n=P^(2q)
Sorry. It's n=(p^2) * q.
then answer is D
Join the discussion