Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

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

For a positive integer n, P(n) is denoted as the product of

Expert replies
by Max@Math Revolution » Wed Sep 25, 2019 11:24 pm
[GMAT math practice question]

For a positive integer n, P(n) is denoted as the product of each digit of n. For example, P(29)=2*9=18, P(457)=4*5*7=140. a, b and c are 2-digit positive integers with P(a)P(b)P(c)=9. What is the maximum value of a+b+c?

A. 91
B. 102
C. 113
D. 124
E. 135
Join the discussion
Source: — Problem Solving |

by deloitte247 » Fri Sep 27, 2019 10:56 pm
P(a) * P(b) * P(c)=9
Factors of a= 1, 3, and 9
Therefore, the possible values of P(a), P(b) and P(c) could be;
P(a) = P(11) = 1*1 = 1
P(b) = P(11) = 1*1 = 1
P(c) = P(91) = 9*1 = 9
===> P(a) * P(b) * P(c) = 1*1*9 = 9
OR
P(a) = P(11) = 1*1 = 1
P(b) = P(31) = 3*1 = 3
P(c) = P(31) = 3*1 = 3
===> P(a) * P(b) * P(c) = 1*3*3 = 9
OR
P(a) = P(33) = 3*3 = 9
P(b) = P(11) = 1*1 = 1
P(c) = P(11) = 1*1 = 1
===> P(a) * P(b) * P(c) = 9*1*1 = 9
OR
P(a) = P(11) = 1*1 = 1
P(b) = P(19) = 1*9 = 9
P(c) = P(11) = 1*1 = 1
===> P(a) * P(b) * P(c) = 1*9*1 = 9

Out of all these possible values, the maximum value of (a+b+c) will occur when P(a)=P(11), P(b)=P(11), and P(c)=P(91), thereby making a=11; b=11 and c=91.
Therefore, (a+b+c) = 11 +11 +91 = 113

Option C is the correct answer.

Hope this helps? Thanks
Join the discussion

by Max@Math Revolution » Sun Sep 29, 2019 5:18 pm
=>

In order to have a maximum value of P(a)P(b)P(c), the possible cases of tuples (a,b,c) we should consider are (91,11,11) and (31,31,11).
For (91,11,11), we have 91 + 11 + 11 = 113.
For (31,31,11), we have 31 + 31 + 11 = 73.
Thus, the maximum of a + b + c is 113.

Therefore, C is the answer.
Answer: C
Join the discussion