Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 19 Nov 2007
Posts: 223

The number 523abc is divisible by 7,8,9. Then what is the [#permalink]
Show Tags
06 Nov 2009, 03:41
22
This post was BOOKMARKED
Question Stats:
34% (04:05) correct
66% (02:12) wrong based on 290 sessions
HideShow timer Statistics
The number 523abc is divisible by 7,8,9. Then what is the value of a*b*c A. 504 B. 532 C. 210 D. 180 E. 280
Official Answer and Stats are available only to registered users. Register/ Login.
Last edited by Bunuel on 09 Jul 2013, 10:44, edited 1 time in total.
Renamed the topic and edited the question.



Intern
Affiliations: CA  India
Joined: 27 Oct 2009
Posts: 45
Location: India
Schools: ISB  Hyderabad, NSU  Singapore

Re: The number 523abc [#permalink]
Show Tags
06 Nov 2009, 05:45
6
This post received KUDOS
I dont know if my method is the best, but i guess the ans is surely available and quicker as well. Here i go:
523abc is divisible by 7, 8 and 9. product of 7, 8 and 9 is 504 and hence 523abc must be divisible by 504 as well. Lets assume, abc are 000, dividing 523000 by 504, we get a remainder of 352, which means it is short by 152 (504352) to be perfectly divisible by 504. so the number has to be 523152 or 523656 (which is 523152+504). so abc should be 1,5,2 or 6,5,6. Product of these sets of integers is, 10 or 180. Since 10 is not in the option, 180 should be the answer.
I would go with OA D.



Math Expert
Joined: 02 Sep 2009
Posts: 39702

The number 523abc is divisible by 7,8,9. Then what is the [#permalink]
Show Tags
06 Nov 2009, 06:14
9
This post received KUDOS
Expert's post
9
This post was BOOKMARKED



Intern
Joined: 30 Dec 2008
Posts: 21

Re: The number 523abc [#permalink]
Show Tags
06 Nov 2009, 08:21
I also choose D. a, b, c cannot be 0. The number is divisible by 9 =>sum of 5+2+3+a+b+c must be divisible by 9, so a+b+c can be 8, 17 or 26 (as maximum sum of 3 1digit number is 27). It is also divisible by 8 => It will be divisible by 4 and 2 => bc must be divisible by 4 and c must be even. so c just can be 2 or 6. It is also divisible by 7. So abc must be 656, its product is 180.



VP
Status: Current Student
Joined: 24 Aug 2010
Posts: 1345
Location: United States
WE: Sales (Consumer Products)

Re: The number 523abc [#permalink]
Show Tags
09 May 2011, 18:59
I broke the answer choices into their prime factors, then tested the different 3 digit combos ending in an even number for which one would make 5+2+3+a+b+c be divisible by 9. The only number with prime factors that does that is 180. Hence, the answer is D.
_________________
The Brain Dump  From Low GPA to Top MBA (Updated September 1, 2013)  A Few of My Favorite Things> http://cheetarah1980.blogspot.com



Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance

Re: The number 523abc [#permalink]
Show Tags
07 Jan 2014, 11:09
Bunuel wrote: jade3 wrote: The number 523abc is divisible by 7,8,9. Then what is the value of a*b*c
A. 504 B. 532 C. 210 D. 180 E. 280 LCM of 7, 8 and 9 is 504, thus 523abc must be divisible by 504. 523abc=523000+abc 523000 divided by 504 gives a remainder of 352. Hence, 352+abc=k*504. k=1 abc=152 > a*b*c=10 k=2 abc=656 > a*b*c=180 As abc is three digit number k can not be more than 2. Two answers? Well only one is listed in answer choices, so D. Answer: D. I thought that for a number to be divisible by 8, its last three digits had to be divisible by 8, any clue on why this is not working in answer choice D? Cheers! J



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7443
Location: Pune, India

Re: The number 523abc [#permalink]
Show Tags
07 Jan 2014, 20:44
jlgdr wrote: Bunuel wrote: jade3 wrote: The number 523abc is divisible by 7,8,9. Then what is the value of a*b*c
A. 504 B. 532 C. 210 D. 180 E. 280 LCM of 7, 8 and 9 is 504, thus 523abc must be divisible by 504. 523abc=523000+abc 523000 divided by 504 gives a remainder of 352. Hence, 352+abc=k*504. k=1 abc=152 > a*b*c=10 k=2 abc=656 > a*b*c=180 As abc is three digit number k can not be more than 2. Two answers? Well only one is listed in answer choices, so D. Answer: D. I thought that for a number to be divisible by 8, its last three digits had to be divisible by 8, any clue on why this is not working in answer choice D? Cheers! J The options give us the value of a*b*c, not abc. If we get abc as 656, it is divisible by 8. The options give us a*b*c = 6*5*6 = 180. The product may or may not be divisible by 8.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance

Re: The number 523abc [#permalink]
Show Tags
12 Feb 2014, 07:53
Thanks Karishma, anyways what I tried here was prime factorization to see which answer choice could be multiples of 9, that is having the whole sum of digits to be a multiple of 9 and only C and D worked. Now, it seems a bit complicated to decide between the two. I have both even too. Any clues on how to proceed from here? I still haven't used the fact that they need to be divisible by 7 and 8 but I'm not sure how to approach these tests quickly
Kudos for a good reply! Cheers J



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7443
Location: Pune, India

Re: The number 523abc [#permalink]
Show Tags
12 Feb 2014, 21:01
7
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
jlgdr wrote: Thanks Karishma, anyways what I tried here was prime factorization to see which answer choice could be multiples of 9, that is having the whole sum of digits to be a multiple of 9 and only C and D worked. Now, it seems a bit complicated to decide between the two. I have both even too. Any clues on how to proceed from here? I still haven't used the fact that they need to be divisible by 7 and 8 but I'm not sure how to approach these tests quickly
Kudos for a good reply! Cheers J 523abc is divisible by 9 which means 5+2+3+a+b+c is divisible by 9 i.e. 1+a+b+c is divisible by 9. When you prime factorize each option, you are left only with (D) A. \(504 = 2^3 * 3^2 * 7\) Since a, b and c must be single digit numbers, the only way you can assign values to them is 8, 9 and 7. 1+8+9+7 = 25 which is not divisible by 9. B. \(532 = 2^2*7*19\) 19 cannot be assigned to any of a, b and c and hence this product is not possible. C. \(210 = 2^2 * 3 * 5 * 7\) You cannot combine them in any way such that you get three single digit numbers. Note that if 5 and 7 are multiplied by even 2, they will become double digit hence two of a, b and c must take values of 5 and 7. The third cannot be the leftover 12. D. \(180 = 2^2 * 3^2 * 5\) There are two options: 4, 9, 5 and 6, 6, 5. 6, 6, 5 satisfies since 1+6+6+5 = 18 divisible by 9 E. \(280 = 2^3 * 5 * 7\) The digits must be 8, 5 and 7. But 1+8+5+7 is not divisible by 9. Answer (D)
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Manager
Joined: 24 Oct 2013
Posts: 174
Location: Canada
WE: Design (Transportation)

Re: The number 523abc [#permalink]
Show Tags
20 Apr 2014, 11:28
Bunuel wrote: jade3 wrote: The number 523abc is divisible by 7,8,9. Then what is the value of a*b*c
A. 504 B. 532 C. 210 D. 180 E. 280 LCM of 7, 8 and 9 is 504, thus 523abc must be divisible by 504. 523abc=523000+abc. 523000 divided by 504 gives a remainder of 352. Hence, 352+abc=k*504. k=1 abc=152 > a*b*c=10. k=2 abc=656 > a*b*c=180. As abc is three digit number k can not be more than 2. Two answers? Well only one is listed in answer choices, so D. Answer: D. I have a doubt Bunuel. Wouldn't that be equal to aba? because i tried solving it backwards using options. The only way option D could've factorized into single digits was 4,5,9 and for this case the singular digits would add up to 5+2+3+4+5+9 = 28 which would make it indivisible by 9. i understand it could have been 6,6,5 too but wouldn't that make it aba instead of abc?



Math Expert
Joined: 02 Sep 2009
Posts: 39702

Re: The number 523abc [#permalink]
Show Tags
20 Apr 2014, 12:05
gauravkaushik8591 wrote: Bunuel wrote: jade3 wrote: The number 523abc is divisible by 7,8,9. Then what is the value of a*b*c
A. 504 B. 532 C. 210 D. 180 E. 280 LCM of 7, 8 and 9 is 504, thus 523abc must be divisible by 504. 523abc=523000+abc. 523000 divided by 504 gives a remainder of 352. Hence, 352+abc=k*504. k=1 abc=152 > a*b*c=10. k=2 abc=656 > a*b*c=180. As abc is three digit number k can not be more than 2. Two answers? Well only one is listed in answer choices, so D. Answer: D. I have a doubt Bunuel. Wouldn't that be equal to aba? because i tried solving it backwards using options. The only way option D could've factorized into single digits was 4,5,9 and for this case the singular digits would add up to 5+2+3+4+5+9 = 28 which would make it indivisible by 9. i understand it could have been 6,6,5 too but wouldn't that make it aba instead of abc? I think you assume that a, b and c must be distinct digits. But this is not true: unless it is explicitly stated otherwise, different variables CAN represent the same number. Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 30 May 2013
Posts: 188
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82

Re: The number 523abc [#permalink]
Show Tags
22 May 2014, 11:01
VeritasPrepKarishma wrote: jlgdr wrote: Thanks Karishma, anyways what I tried here was prime factorization to see which answer choice could be multiples of 9, that is having the whole sum of digits to be a multiple of 9 and only C and D worked. Now, it seems a bit complicated to decide between the two. I have both even too. Any clues on how to proceed from here? I still haven't used the fact that they need to be divisible by 7 and 8 but I'm not sure how to approach these tests quickly
Kudos for a good reply! Cheers J 523abc is divisible by 9 which means 5+2+3+a+b+c is divisible by 9 i.e. 1+a+b+c is divisible by 9. When you prime factorize each option, you are left only with (D) A. \(504 = 2^3 * 3^2 * 7\) Since a, b and c must be single digit numbers, the only way you can assign values to them is 8, 9 and 7. 1+8+9+7 = 25 which is not divisible by 9. B. \(532 = 2^2*7*19\) 19 cannot be assigned to any of a, b and c and hence this product is not possible. C. \(210 = 2^2 * 3 * 5 * 7\) You cannot combine them in any way such that you get three single digit numbers. Note that if 5 and 7 are multiplied by even 2, they will become double digit hence two of a, b and c must take values of 5 and 7. The third cannot be the leftover 12. D. \(180 = 2^2 * 3^2 * 5\) There are two options: 4, 9, 5 and 6, 6, 5. 6, 6, 5 satisfies since 1+6+6+5 = 18 divisible by 9 E. \(280 = 2^3 * 5 * 7\) The digits must be 8, 5 and 7. But 1+8+5+7 is not divisible by 9. Answer (D) I find this method to be very time consuming.



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1857
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: The number 523abc is divisible by 7,8,9. Then what is the [#permalink]
Show Tags
22 May 2014, 22:22
LCM of 7,8 & 9 = 7 * 8 * 9 = 504 Whatever the number 523abc should be divisible by 504 We know that 504000 is divisible by 504 Adding multiples of 504 to 504000 to reach 523abc We get two numbers: 523 152 & 523 656 (both are divisible by 504) Product of abc = 10 OR 180 Option D fits in Answer = 180 = D
_________________
Kindly press "+1 Kudos" to appreciate



Manager
Joined: 30 May 2013
Posts: 188
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82

Re: The number 523abc is divisible by 7,8,9. Then what is the [#permalink]
Show Tags
23 May 2014, 07:23
523abc is divisible by 504.
To find abc first lets divide 523000 by 504
523000/504 = 137 and remainder is 352. (1)
352 + 152 = 504 i.e. when 152 is added with the remainder 352 of (1) then in (1) quotient = 138 and remainder will be 0
523152 / 504 = 138 and remainder is 0 As per the question "abc" is a*b*c = 180 but here 1*5*2=10
523656/504 = 139 and remainder is 0 a*b*c= 6*5*6 = 180
Regards, Swami.



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15990

Re: The number 523abc is divisible by 7,8,9. Then what is the [#permalink]
Show Tags
25 May 2015, 05:57
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2185

Re: The number 523abc is divisible by 7,8,9. Then what is the [#permalink]
Show Tags
14 Mar 2016, 02:44
PareshGmat wrote: LCM of 7,8 & 9 = 7 * 8 * 9 = 504
Whatever the number 523abc should be divisible by 504
We know that 504000 is divisible by 504
Adding multiples of 504 to 504000 to reach 523abc
We get two numbers:
523152 & 523656 (both are divisible by 504)
Product of abc = 10 OR 180
Option D fits in
Answer = 180 = D What ? Okay so you are telling me to find this hepttyy looking divisibility computation.. Looks Wrong to me.. answer is not doubt right but the method is tooooooooooo long P.S => add a few more ooooo in tooooo
_________________
Give me a hell yeah ...!!!!!



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15990

Re: The number 523abc is divisible by 7,8,9. Then what is the [#permalink]
Show Tags
09 Apr 2017, 08:24
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: The number 523abc is divisible by 7,8,9. Then what is the
[#permalink]
09 Apr 2017, 08:24







