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If x & y are consecutive positive integer multiples of

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CEO
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If x & y are consecutive positive integer multiples of [#permalink] New post 11 Sep 2003, 03:55
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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
CEO
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 [#permalink] New post 11 Sep 2003, 05:36
wonder_gmat wrote:
18


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

thanks
praetorian
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 [#permalink] New post 11 Sep 2003, 09:33
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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approach [#permalink] New post 11 Sep 2003, 22:28
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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 [#permalink] New post 11 Sep 2003, 22:46
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
  [#permalink] 11 Sep 2003, 22:46
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