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by cveluswamy » Mon Aug 06, 2007 6:51 pm
Pls Explain

Statement 1: n = multiple of 20 is 20/15 an Int?

Not Sufficient.

Statement 2: n + 6 multiple of 3 say n = 3 which is 3/15 is 3/15 an int?

Not sufficient.

Together: 20 + 6/3 is an int? NO.. But the OA is C.. Can some one explain. Thanks
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by sochatte » Mon Aug 06, 2007 8:40 pm
If n is multiple of 15 then it can be
n = m(any number) x 15 = m x 5 x 3

1. n is a multiple of 20
that means n = a(any number) x 5 x 2 x 2
here if a is 3 , n will be multiple of 20.However, we are not sure.
So not sufficient.

2. n + 6 is a multiple of 3
This statement tells us that 3 is one of the factor of n. However, we don't know whether 5 is a factor of n.

Combining both the statements we know that 5(from 1) and 3 (from2) both are factors of n.Hence C is the correct ans.
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by cveluswamy » Mon Aug 06, 2007 9:25 pm
Thanks. That helps a lot.
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by bingojohn » Tue Aug 07, 2007 1:42 pm
sochatte wrote:If n is multiple of 15 then it can be
n = m(any number) x 15 = m x 5 x 3

1. n is a multiple of 20
that means n = a(any number) x 5 x 2 x 2
here if a is 3 , n will be multiple of 20.However, we are not sure.
So not sufficient.

2. n + 6 is a multiple of 3
This statement tells us that 3 is one of the factor of n. However, we don't know whether 5 is a factor of n.

Combining both the statements we know that 5(from 1) and 3 (from2) both are factors of n.Hence C is the correct ans.
Just wanted to elaborate a little further that the trick really was to decipher statement (2) properly. (n+6) is a multiple of 3 can also be written in equation form as

(n+6) = 3(k), where k is a integer
=> n + 3(2) = 3(k)
=> n = 3(k) - 3(2)
=> n = 3 (k-2)
hence n is a multiple of 3... and then the rest follows...
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