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 Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget 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

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

number properties

Expert replies
by sud21 » Sat Jan 28, 2012 11:14 pm
the unit' digit of a number is twice the hundredth' digit. And the number is divisible by 3. Which of the following must be true?
I. the tenth digit is nonzero
II. the tenth digit is odd
III. the tenth digit is a multiple of 3
Join the discussion
Source: — Problem Solving |

by rijul007 » Sat Jan 28, 2012 11:26 pm
sud21 wrote:the unit' digit of a number is twice the hundredth' digit. And the number is divisible by 3. Which of the following must be true?
I. the tenth digit is nonzero
II. the tenth digit is odd
III. the tenth digit is a multiple of 3
I. the tenth digit is nonzero
consider the number 603
this statement cannot be always correct


II. the tenth digit is odd
603, 864,..

not always true

III. the tenth digit is a multiple of 3
let us say units digit is x
hundredth digit is 2x
tenth digit is a
the number is multiple of 3
hence, 2x+a+x = 3n
a = 3(n-x)

tenth digit is a multiple of 3

True

Option III is correct
Join the discussion

by Abhishek009 » Sun Jan 29, 2012 10:03 am
sud21 wrote:the unit' digit of a number is twice the hundredth' digit. And the number is divisible by 3. Which of the following must be true?
I. the tenth digit is nonzero
II. the tenth digit is odd
III. the tenth digit is a multiple of 3
The structure of the number will b as follows -

H T U

Now it's given -

The Unit' digit of a number is twice the hundredth' digit..


So , the digits will be -

H T 2H

Summation of the digits will be -

3H + T

Now in order to make the number divisible by 3 T must be a multiple of 3 , or we can write T as a multiple of 3 ie , 3k


So now the summation of the number stands as follows -

3H + 3k



Now let's consider the problem statements given -

I. the tenth digit is nonzero

Now the units digit can't b 0 , coz then the Hundredth's digit must have to b zero and it's not possible ...

II. the tenth digit is odd

Now this can not be possible , coz anything multiplied by 2 makes it an even number...

III. the tenth digit is a multiple of 3

Now this has just been proved above , so it's true...
Abhishek
Join the discussion

by ronnie1985 » Sun Jan 29, 2012 10:08 am
Let the no be "xy(2x)"
The no is then 100x+y+2x = 102x+y which is divisible by 3. Thus, 3x+y is also a multiple of 3, which implies y must be a multiple of 3.
Follow your passion, Success as perceived by others shall follow you
Join the discussion