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Source: — Data Sufficiency |
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by parallel_chase » Sat Aug 09, 2008 11:09 am
From the question stem we have to find if n is a multiple of 15 i.e. 5 and 3

If we know that n is a multiple of 5 and 3 both n will also be a multiple of 15.

Statement I

n is a multiple of 20

20= 2*5*2

This means that n will be a multiple of 5 for sure, but we dont know anything about 3. Therefore insufficient.

Statement II
n+6 is a multiple of 3 this means that n is a multiple of 3, but we dont know anything about 5. Therefore insufficient.


Statement I & II

n is a multiple of 5, n is multiple of 3, therefore n has to be a multiple of 15.

Here is rule that you can learn. It is regarding how I found that n will be a multiple of 3 in II statement n+6 is a multiple of 3

If a and b are divisible by n then a-b & a+b both will be divisible by n.

Let me know if you have any doubts.
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by raunekk » Sat Aug 09, 2008 11:52 am
@parallel chase
n is a multiple of 5, n is multiple of 3, therefore n has to be a multiple of 15.

Here is rule that you can learn. It is regarding how I found that n will be a multiple of 3 in II statement n+6 is a multiple of 3

If a and b are divisible by n then a-b & a+b both will be divisible by n.
can you please elaborate this,,,,Even i choose E.

Thank you.
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by parallel_chase » Sat Aug 09, 2008 12:07 pm
If a and b are divisible by n then a-b & a+b both will be divisible by n.

Explanation:

lets say a= 5, b = 25, then a-b & a+b will always be a multiple of 5

5-25 = -20 is a multiple of 5
5+25 = 30 is a multiple of 5


Lets say a=28, b=14 both are multiples of 2 and 7
Therefore,

28-14 = 14 is a multiple of 2 and 7
28+14 = 42 is a multiple of 2 and 7

Now coming to the statement II

Lets say n+6 = X

X is a multiple of 3

6 is a multiple of 3
X is a multiple of 3

Therefore n has to be a multiple of 3

Hope its clear. Let me know if you still have any questions.
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