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
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
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.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

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

GMAT Prep? (Function) Pract 2

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 446
Joined: Thu Jul 26, 2007 1:07 pm
Thanked: 6 times
by dferm » Tue May 13, 2008 3:36 pm
The function f is defined for all positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is any prime number then f(p) =

A. p-1
B. p-2
C. p+1/2
D. p-1/2
E. 2

Please Explain...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 192
Joined: Tue Apr 29, 2008 7:41 am
Thanked: 14 times

by aatech » Wed May 14, 2008 12:46 pm
I will go with A...

IMO, best approach will be to calculate f(n) for few nos

f(2) = 1 => 1 no

f(3) = 2, 1 => 2 nos

f(5) = 4, 3, 2, 1 => 4 nos

f(7) = 6, 5, 4, 3, 2, 1 => 6 nos

and A satisfies the condition for all prime nos

OA please??
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 446
Joined: Thu Jul 26, 2007 1:07 pm
Thanked: 6 times

by dferm » Wed May 14, 2008 1:57 pm
The answer is A but don't quite understand this question..

Can you please put a little more light..


Thanks..

PS. ur the best
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 192
Joined: Tue Apr 29, 2008 7:41 am
Thanked: 14 times

by aatech » Thu May 15, 2008 5:38 am
Thanks for the compliments...

I will try my best to explain.. hope I am able to do so...

the question says f(n) is NUMBER of positive integers those are less than n and these numbers do
not have any common factors with n itself..

So, for any +ve integer n the function will contain all numbers less than n.. Since question is
concerned about only prime numbers you should not worry about the latter part as, for sure, there
will not be any common factors for prime numbers...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1