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# If x & y are consecutive positive integer multiples of

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CEO
Joined: 15 Aug 2003
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If x & y are consecutive positive integer multiples of [#permalink]

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11 Sep 2003, 04:55
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4.If x & y are consecutive positive integer multiples of 3, what is the greatest integer j such that (xy) / j is always an integer?

A.27

B.18

C.9

D.6

E.3

Kudos [?]: 919 [0], given: 781

CEO
Joined: 15 Aug 2003
Posts: 3454

Kudos [?]: 919 [0], given: 781

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11 Sep 2003, 06:36
wonder_gmat wrote:
18

I am more interested in the approach... mind explaining the answer?

thanks
praetorian

Kudos [?]: 919 [0], given: 781

Senior Manager
Joined: 22 May 2003
Posts: 328

Kudos [?]: 173 [0], given: 0

Location: Uruguay

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11 Sep 2003, 10:33
I agree 18

Each x and y are divisible by 3. Since they are consecutive multiples of 3 one of them is even (divisible by 2).

So x*y is divisible by 3*3*2=18

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Intern
Joined: 18 Aug 2003
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Location: USA

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11 Sep 2003, 23:28
since x and y are consecutive multiples of 3, naturally one of the two is a multiple of 6. so ur denominator, j has to be a multiple of 6. since 18 is 6x3 one of x/y will be divisible by 6 and the other by 3.
hence 18.

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CEO
Joined: 15 Aug 2003
Posts: 3454

Kudos [?]: 919 [0], given: 781

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11 Sep 2003, 23:46
MartinMag wrote:
I agree 18

Each x and y are divisible by 3. Since they are consecutive multiples of 3 one of them is even (divisible by 2).
So x*y is divisible by 3*3*2=18

You guys are good at this kind of thing...i happen to solve it this way..

x = 3* m ( m=1,2,3,4,5,6)
y = 3 * n ( n=2,3,4,5)

Therefore xy /j = 9 mn /j ..

Any value of j MUST divide the smallest value of 9mn..which we get by plugging in the minimum possible value of m and n....that is.. m= 1, n =2.

We get ...18/j and the largest integer that can divide this is 18..

Think this is ok ?

Thanks
Praetorian

Kudos [?]: 919 [0], given: 781

11 Sep 2003, 23:46
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# If x & y are consecutive positive integer multiples of

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