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The greatest number of four digits which is divisible by 15, 25, 40 an

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The greatest number of four digits which is divisible by 15, 25, 40 an  [#permalink]

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New post Updated on: 09 Aug 2018, 23:19
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The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000
B. 9400
C. 9600
D. 9800
E. 9999

Originally posted by LMP on 09 Aug 2018, 01:53.
Last edited by Bunuel on 09 Aug 2018, 23:19, edited 1 time in total.
Renamed the topic.
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Re: The greatest number of four digits which is divisible by 15, 25, 40 an  [#permalink]

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New post 09 Aug 2018, 02:20
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Bulusuchaitanya wrote:
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000
B. 9400
C. 9600
D. 9800
E. 9999


Since the 4-digit number is divisible by 15, 25, 40 and 75. So, the number must be divisible by LCM( 15, 25, 40 ,75)=600.

The greatest 4-digit number in number system is 9999.

So, the required number=9999-Rem(\(\frac{9999}{600}\))=9999-399=9600

Ans. (C)
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Re: The greatest number of four digits which is divisible by 15, 25, 40 an  [#permalink]

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New post 09 Aug 2018, 03:51
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Bulusuchaitanya wrote:
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000
B. 9400
C. 9600
D. 9800
E. 9999



the choices straight away give you the answer..
the number has to be multiple of 5 and 2, so has to have 0 as units digit... eliminate E
the number is also a multiple of 3, so the sum of digits of number must be divisible by 3...
9800=9+8+0+0=17.. NOT divisible by 3 so eliminate D
9600 = 9+6+0+0=15...div by 3, so our answer
C

Also the question is basically asking for multiple of the LCM of 15,25,40 and 75
\(15=3*5...........
25=5^2...............
40=2^3*5...........
75=3*5^2\)
LCM = \(2^3*3*5^2=600\)
in choices 9600 is the largest multiple of 6...
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Re: The greatest number of four digits which is divisible by 15, 25, 40 an  [#permalink]

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New post 29 Oct 2018, 16:17
chetan2u wrote:
Bulusuchaitanya wrote:
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000
B. 9400
C. 9600
D. 9800
E. 9999



the choices straight away give you the answer..
the number has to be multiple of 5 and 2, so has to have 0 as units digit... eliminate E
the number is also a multiple of 3, so the sum of digits of number must be divisible by 3...
9800=9+8+0+0=17.. NOT divisible by 3 so eliminate D
9600 = 9+6+0+0=15...div by 3, so our answer
C

Also the question is basically asking for multiple of the LCM of 15,25,40 and 75
\(15=3*5...........
25=5^2...............
40=2^3*5...........
75=3*5^2\)
LCM = \(2^3*3*5^2=600\)
in choices 9600 is the largest multiple of 6...


I solved this by looking at the GCF ... which is 15 in this case.....
then i looked at each number which was divisible by 15 ... turned out to be C.

I wonder if my logic is correct?
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Re: The greatest number of four digits which is divisible by 15, 25, 40 an  [#permalink]

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New post 29 Oct 2018, 16:42
Bulusuchaitanya wrote:
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000
B. 9400
C. 9600
D. 9800
E. 9999


I would solve this by elimination.

B, D are not multiples of 3 whereas 15, 75 are. So these can be eliminated.

E is odd.

A is too small so I would test 9600 and I see that it is divisible by all.

C it is.
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Re: The greatest number of four digits which is divisible by 15, 25, 40 an  [#permalink]

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New post 30 Oct 2018, 18:07
Bulusuchaitanya wrote:
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000
B. 9400
C. 9600
D. 9800
E. 9999


Let’s determine the LCM of 15, 25, 40, and 75:

15 = 3 x 5

25 = 5^2

40 = 2^3 x 5

75 = 5^2 x 3

So the LCM is 2^3 x 3 x 5^2 = 8 x 3 x 25 = 600, and the largest 4-digit number that is a multiple of 600 is 9600.

Answer: C
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Re: The greatest number of four digits which is divisible by 15, 25, 40 an &nbs [#permalink] 30 Oct 2018, 18:07
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