Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 04 Sep 2015, 11:56

### 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The number 523abc is divisible by 7,8,9. Then what is the

Author Message
TAGS:
Manager
Joined: 19 Nov 2007
Posts: 227
Followers: 1

Kudos [?]: 149 [1] , given: 1

The number 523abc is divisible by 7,8,9. Then what is the [#permalink]  06 Nov 2009, 02:41
1
KUDOS
14
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

34% (04:19) correct 66% (02:12) wrong based on 188 sessions
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
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Jul 2013, 09:44, 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
Followers: 16

Kudos [?]: 546 [2] , given: 5

Re: The number 523abc [#permalink]  06 Nov 2009, 04:45
2
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 (504-352) 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: 29210
Followers: 4753

Kudos [?]: 50354 [6] , given: 7544

The number 523abc is divisible by 7,8,9. Then what is the [#permalink]  06 Nov 2009, 05:14
6
KUDOS
Expert's post
5
This post was
BOOKMARKED
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 cannot be more than 2.

Two answers? Well only one is listed in answer choices, so D.

_________________
Intern
Joined: 30 Dec 2008
Posts: 21
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: The number 523abc [#permalink]  06 Nov 2009, 07:21
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 1-digit 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
GMAT 1: 710 Q48 V40
WE: Sales (Consumer Products)
Followers: 105

Kudos [?]: 412 [0], given: 73

Re: The number 523abc [#permalink]  09 May 2011, 17:59
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

SVP
Joined: 06 Sep 2013
Posts: 2046
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 32

Kudos [?]: 346 [0], given: 355

Re: The number 523abc [#permalink]  07 Jan 2014, 10:09
Bunuel 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.

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: 5870
Location: Pune, India
Followers: 1485

Kudos [?]: 8012 [0], given: 190

Re: The number 523abc [#permalink]  07 Jan 2014, 19:44
Expert's post
jlgdr wrote:
Bunuel 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.

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 SVP Joined: 06 Sep 2013 Posts: 2046 Concentration: Finance GMAT 1: 770 Q0 V Followers: 32 Kudos [?]: 346 [0], given: 355 Re: The number 523abc [#permalink] 12 Feb 2014, 06:53 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: 5870 Location: Pune, India Followers: 1485 Kudos [?]: 8012 [5] , given: 190 Re: The number 523abc [#permalink] 12 Feb 2014, 20:01 5 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: 178
GMAT 1: 720 Q49 V38
WE: Design (Transportation)
Followers: 6

Kudos [?]: 11 [0], given: 82

Re: The number 523abc [#permalink]  20 Apr 2014, 10:28
Bunuel 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.

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: 29210
Followers: 4753

Kudos [?]: 50354 [0], given: 7544

Re: The number 523abc [#permalink]  20 Apr 2014, 11:05
Expert's post
gauravkaushik8591 wrote:
Bunuel 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.

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.
_________________
Manager
Joined: 30 May 2013
Posts: 193
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Followers: 0

Kudos [?]: 40 [0], given: 72

Re: The number 523abc [#permalink]  22 May 2014, 10:01
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

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.

I find this method to be very time consuming.
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1859
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 26

Kudos [?]: 1074 [0], given: 193

Re: The number 523abc is divisible by 7,8,9. Then what is the [#permalink]  22 May 2014, 21:22
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:

523152 & 523656 (both are divisible by 504)

Product of abc = 10 OR 180

Option D fits in

_________________

Kindly press "+1 Kudos" to appreciate

Manager
Joined: 30 May 2013
Posts: 193
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Followers: 0

Kudos [?]: 40 [0], given: 72

Re: The number 523abc is divisible by 7,8,9. Then what is the [#permalink]  23 May 2014, 06:23
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: 6218
Followers: 346

Kudos [?]: 71 [0], given: 0

Re: The number 523abc is divisible by 7,8,9. Then what is the [#permalink]  25 May 2015, 04:57
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.
_________________
Re: The number 523abc is divisible by 7,8,9. Then what is the   [#permalink] 25 May 2015, 04:57
Similar topics Replies Last post
Similar
Topics:
3 The remainder of the division of a number by 63 is 27. What 5 05 Aug 2014, 08:24
8 If y^4 is divisible by 60, what is the minimum number of dis 15 06 Jan 2013, 06:50
A number 15B is divisible by 6 3 14 Nov 2012, 07:37
6 If the number x3458623y is divisible by 88, what is the 6 08 Jun 2012, 08:21
If a number (N) is divisible by 33, what will be the minimum 7 08 Jun 2012, 08:20
Display posts from previous: Sort by