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
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.

Number properties + probability

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 57
Joined: Sat Jan 24, 2009 7:45 am
Thanked: 1 times
by ellexay » Wed Feb 11, 2009 7:23 am
) If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8?



Soln: Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000
the first one is 8 and the last one is 1000
==> ((1000-8)/8) + 1 = 125
==> p (picking a multiple of 2 & 8) = 125/1000 = 1/8

Where did they get this 1000-8/8 +1 formula from? Can someone verify this answer? Thank you.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 amFollowed by: 550 members
Thanked: 712 times
GMAT Score:770

by DanaJ » Wed Feb 11, 2009 7:57 am
I personally have no idea where they took that formula from. I know what i would do: divide 1000 by 8 and, after noticing that 1000 is a multiple of 8, then it is clear that the number of multiples of 8 is 1000/8 = 125. If you had smth that was not a multiple of 8 (say 1003), I'd divide this number by 8 as well and use the quotient for my answer.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 258
Joined: Thu Aug 07, 2008 5:32 am
Thanked: 16 times

Re: Number properties + probability

by x2suresh » Wed Feb 11, 2009 9:25 am
ellexay wrote:) If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8?



Soln: Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000
the first one is 8 and the last one is 1000
==> ((1000-8)/8) + 1 = 125
==> p (picking a multiple of 2 & 8) = 125/1000 = 1/8

Where did they get this 1000-8/8 +1 formula from? Can someone verify this answer? Thank you.

first multiple of 8 between 1 and 1000 =8
last multiple of 8 between 1 and 1000 = 1000

total number of multiples = (1000-8)/8 = 124
Above forumula excluding number 8 itself..
So final answer = 124+1=125
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 424
Joined: Sun Dec 07, 2008 5:15 pm
Location: Sydney
Thanked: 12 times

Re: Number properties + probability

by piyush_nitt » Thu Feb 12, 2009 4:36 am
ellexay wrote:) If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8?



Soln: Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000
the first one is 8 and the last one is 1000
==> ((1000-8)/8) + 1 = 125
==> p (picking a multiple of 2 & 8) = 125/1000 = 1/8

Where did they get this 1000-8/8 +1 formula from? Can someone verify this answer? Thank you.
AP formula

to determine the nth term

An = A+ (n-1)d

where An - nth Term
A - first term
n - number of terms
d - difference

In this case

1000 = 8 + (n-1)8

(1000-8) = (n-1)8

(1000-8)/8 + 1 = n
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1