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Re: The number 523abc [#permalink]
06 Nov 2009, 04:45

1

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 (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.

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.

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

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

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?

Re: The number 523abc [#permalink]
07 Jan 2014, 19:44

Expert's post

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. _________________

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

Re: The number 523abc [#permalink]
12 Feb 2014, 20:01

5

This post received KUDOS

Expert's post

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.

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

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? _________________

CAT1-520 (GMAT CLUB) CAT2-620 (MGMAT 1) CAT3-630 (KAPLAN) CAT4-680 (MGMAT 2) CAT5-710 (VERITAS 1) CAT6-680 (MGMAT 3) CAT7-720 (VERITAS 2) CAT8-730 (VERITAS 3) ... Lets see how far can i stretch it up to.

Re: The number 523abc [#permalink]
20 Apr 2014, 11:05

Expert's post

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.

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

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.