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.

can't understand how to solve

Expert replies
Source: — Problem Solving |

by ashis979 » Sun Oct 04, 2009 11:13 am
27 integers. I'd go with a sets approach here.

T: # of integers between 100 and 150 incl. = 51
A: # of integers between 100 and 150 incl. that are multiples of 3 = # of integers between 34 and 50 incl. (think multiples of 3 between 102 and 150) = 17
B: # of integers between 100 and 150 incl. that are multiples of 5 = # of integers between 20 and 30 incl. = 11
C: Intersection set. Need the number of integers between 100 and 150 incl. that are divisible by both 3 and 5. Essentially, multiples of 15. So that is the same as saying # of integers between 7 and 10 incl. (think multiples of 15 between 105 and 150) = 4

So, # of integers between 100 and 150 incl. that can be evenly divided by neither 3 nor 5 = T-(A+B)+C = 51-28+4=27. [You need to add back the 4 to avaoid double counting, because those 4 are already included in sets A and B).
Join the discussion