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# project GMAT ps

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Manager
Joined: 05 Oct 2005
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project GMAT ps [#permalink]  22 Nov 2005, 15:20
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Question Stats:

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A set consists of all integers between 600 and 1999, inclusive. If a number is selected at random from this set, what is the probability that it is divisible by both 5 and 13?

(A) 65/1399
(B) 13/280
(C) 1/50
(D) 3/200
(E) 21/1399

thanks
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Senior Manager
Joined: 11 May 2004
Posts: 458
Location: New York
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I think its D

1999-600+1 = 1400

So eliminate A and E

I was kind stuck from there, so i started seeing how many numbers are usually divisible by 5 and 13 and i realized it would take ME more than 2 mins, so I just guessed.

Anyone have any thoughts???
Director
Joined: 14 Sep 2005
Posts: 994
Location: South Korea
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Kudos [?]: 44 [0], given: 0

GCD of 5 & 13 = 65

65 * 10 = 650 > 600
65 * 30 = 1950 < 1999

30-10+1=21
1999-600+1=1400

21/1400
= 3/200
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Auge um Auge, Zahn um Zahn !

Current Student
Joined: 29 Jan 2005
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E is the trap answer because 1999-600= 1400 (inclusive) Nix A and E right away.

to be both a multiple of 13 and 5, the number also has to be a multiple of 13*5=65

65*10=650 (lowest common multiple)
65*30=1950 (highest common multiple)

30-10=21 (inclusive)

Probability= #desired outcomes/#possible outcomes

21/1400--->3/200 (D)
Senior Manager
Joined: 03 Nov 2005
Posts: 397
Location: Chicago, IL
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I've done this way. First, I,ve found how many numbers between 600 and 1999 are perfectly divisible by both 5 and 13. Since both 5 and 13 are primes, their LCM is 65. So. there are 600/65=9 and 1999/65=30. 30-9=21 intergers are divisible by 65 between 600 and 1999. number of consecutive intergers between 600 and 1999 is 1999-600+1=1400, 21/1400=3/200. The answer is D.
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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
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Kudos [?]: 184 [0], given: 0

Numer of integers = 1400
Divisible by 5 and 13: must be divisible by 65
1999/65 = 30.75 -> highest multiple of 65 is 30
600/65 = 9.2 --> lowest multiple of 65 is 10

So Total = 30-10+1 = 21
P = 21/1400 = 3/200

Ans: D
Manager
Joined: 05 Oct 2005
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my approach can i do that [#permalink]  23 Nov 2005, 08:28
my approach is
from 60 to 1999 is 1399 but add 1 since inclusive
both 5 AND 13 are prime number so anumber to be divisible by both 13 and 5 must be a multiple of both 5 and 13 HENCE 65

SO I found the number of multiple of 65 in the 14OO numbers
14OO/ 65 = 21 of such number

HENCE final result is 21/1400

By the way i don 't understand why some of you are doing the following :
65*10=650 (lowest common multiple)
65*30=1950 (highest common multiple)

30-10=21 (inclusive)

thanks
_________________

when there is a will there is a way

best regards

SVP
Joined: 28 May 2005
Posts: 1743
Location: Dhaka
Followers: 6

Kudos [?]: 72 [0], given: 0

first of all in order to be divisible by 5 and 13 it needs to be multiple of 65.

65*10 = 650 so we need to start from there.

650*30 = 1950. so between 10 to 30 inclusive we have 21 of them

and from 600 to 1999 there are 1400 numbers.

so probability is 21/1400 = 3/200
_________________

hey ya......

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