Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
25 Nov 2019, 01:50
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
43% (03:10) correct
58%
(02:55)
wrong
based on 160
sessions
History
Date
Time
Result
Not Attempted Yet
ABC is a three-digit number such that ABC = 5(AB + BC + CA), where AB, BC and CA are all two-digit numbers. Find the total number of possible values for the number ABC.
(A) 3
(B) 4
(C) 12
(D) 13
(E) 18
(A) 3
(B) 4
(C) 12
(D) 13
(E) 18
Ansh777
Joined: 03 Nov 2019
Last visit: 06 Jun 2023
Posts: 56
Given Kudos: 129
Location: India
Schools: Booth '24 Columbia CBS '24 Darden '24 Fuqua '24 Haas '24 Harvard '24 LBS '24 Stanford '24 Kellogg '24 Wharton '24
GMAT 1: 710 Q50 V36
Schools: Booth '24 Columbia CBS '24 Darden '24 Fuqua '24 Haas '24 Harvard '24 LBS '24 Stanford '24 Kellogg '24 Wharton '24
GMAT 1: 710 Q50 V36
Posts: 56
26 Nov 2019, 05:33
Kudos
Bookmarks
ABC is a 3-digit no.
So it could be written as
ABC= (100A+10B+C)
Similarly 2-digit numbers
AB= (10A+B)
BC= (10B+C)
CA= (10C+A)
Given that:
100A+10B+C= 5*(10A+B+10B+C+10C+A)
=5*(11A+11B+11C)
=55A+55B+55C
Or
45A-45B=54C
A-B=6C/5
Now since A,B,C are unit, tens or hundreds digit and each of them happen to be first digit(which cannot be zero) of a two digit number ie AB, BC, CA.
So clearly A, B, C lie between 1 & 9
Now again:
A-B=6C/5 ( A-B will also be an integer, hence C needs to be a multiple of 5, Only possible value of C=5)
Now A-B=6
If
A=9 B=3
A=8 B=2
A=7 B=1
A=6 B=0 Not possible.
Therefore only 3 possible values of ABC are: ( 935, 825 & 715)
Answers: A
Posted from my mobile device
So it could be written as
ABC= (100A+10B+C)
Similarly 2-digit numbers
AB= (10A+B)
BC= (10B+C)
CA= (10C+A)
Given that:
100A+10B+C= 5*(10A+B+10B+C+10C+A)
=5*(11A+11B+11C)
=55A+55B+55C
Or
45A-45B=54C
A-B=6C/5
Now since A,B,C are unit, tens or hundreds digit and each of them happen to be first digit(which cannot be zero) of a two digit number ie AB, BC, CA.
So clearly A, B, C lie between 1 & 9
Now again:
A-B=6C/5 ( A-B will also be an integer, hence C needs to be a multiple of 5, Only possible value of C=5)
Now A-B=6
If
A=9 B=3
A=8 B=2
A=7 B=1
A=6 B=0 Not possible.
Therefore only 3 possible values of ABC are: ( 935, 825 & 715)
Answers: A
Posted from my mobile device
26 Nov 2019, 06:09
Kudos
Bookmarks
CaptainLevi
ABC = 5(AB + BC + CA) tells us that ABC is a multiple of 5, so C can be 0 or 5.
But CA will be 2-digit number only when C=5.
Now, AB can be written as 10A+B..
\(5(AB + BC + CA)=5(10A+B+10B+C+10C+A)=5(11(A+B+C))=55(A+B+C)=ABC\)
Thus, ABC is a multiple of 11, and the units digit of A+B should be even, as then only 55(A+B+5) will give AB5..
For AB5 to be multiple of 11, 5+A-B should be multiple of 11...
So
=>5+A-B=0......A-B=-5 ..Discard as A+B is EVEN, which means A-B is also EVEN
or 5+A-B=11......A-B=6..This means least value is when A=7 and B=1 as B cannot be 0
Thus, check for values of A+B=8 onwards
(I) A+B=8
Let us solve
A-B=6 and A+B=8...Add both...2A=14...A=7 and B=1...Number is 715
(II) A+B=10
Let us solve
A-B=6 and A+B=10...Add both...2A=16...A=8 and B=2...Number is 825
(III) A+B=12
Let us solve
A-B=6 and A+B=12...Add both...2A=18...A=9 and B=3...Number is 935
ALL values above 12 can be discarded as it will give A as 2-digit.
for example
A+B=16
Let us solve
A-B=-5 and A+B=16...A and B will come out as fraction..Discard
A-B=6 and A+B=16...Add both...2A=22...A=11 and B=5..Discard
So 3 cases
A
