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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 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.

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

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?

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. _________________

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

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.

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?

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.

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.

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. _________________

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 ...!!!!!

gmatclubot

Re: The number 523abc is divisible by 7,8,9. Then what is the
[#permalink]
14 Mar 2016, 02:44

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

They say you get better at doing something by doing it. then doing it again ... and again ... and again, and you keep doing it until one day you look...