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

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The number 523abc is divisible by 7,8,9. Then what is the [#permalink] New post 06 Nov 2009, 02:41
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A
B
C
D
E

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52% (04:20) correct 47% (02:27) wrong based on 44 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.
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Re: The number 523abc [#permalink] New post 06 Nov 2009, 04:45
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.
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Re: The number 523abc [#permalink] New post 06 Nov 2009, 05:14
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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.
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Re: The number 523abc [#permalink] New post 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.
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Re: The number 523abc [#permalink] New post 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.
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Re: The number 523abc [#permalink] New post 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?

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Re: The number 523abc [#permalink] New post 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.
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Re: The number 523abc [#permalink] New post 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

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