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

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Intern
Joined: 29 Jun 2018
Posts: 16
Location: India
GMAT 1: 530 Q42 V26
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The greatest number of four digits which is divisible by 15, 25, 40 an  [#permalink]

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Updated on: 09 Aug 2018, 23:19
3
00:00

Difficulty:

35% (medium)

Question Stats:

67% (01:13) correct 33% (01:02) wrong based on 78 sessions

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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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09 Aug 2018, 02:20
3
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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09 Aug 2018, 03:51
1
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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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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Posts: 38
Re: The greatest number of four digits which is divisible by 15, 25, 40 an  [#permalink]

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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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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.

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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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