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Is the integer n a mulltiple of 15 ? 1) n is a multiple of

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Is the integer n a mulltiple of 15 ? 1) n is a multiple of [#permalink] New post 30 Jan 2011, 20:39
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88% (01:14) correct 13% (01:37) wrong based on 16 sessions
Is the integer n a mulltiple of 15 ?

1) n is a multiple of 20

2) n + 6 is a multiple of 3


Please explain how to derive the answer to this question. Thanks
[Reveal] Spoiler: OA
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Re: multiple of 15 [#permalink] New post 30 Jan 2011, 21:13
Wayxi wrote:
Is the integer n a multiple of 15 ?

1) n is a multiple of 20

2) n + 6 is a multiple of 3


Please explain how to derive the answer to this question. Thanks


Start with statement 1. It is not sufficient because for a number to be a multiple for 15, it should be divisible by both 3 and 5. Clearly, 20 does not have 3 as a factor. Hence, statement 1 is not sufficient.

Statement 2 says that n+6 is a multiple of three. If you break it down, it is basically saying that n is a multiple of 3. But that we don't know whether n is a multiple of 5 or not. Hence, this is not sufficient.

If you combine both 1 and 2, you can conclude that n has both 5 (from statement 1)and 3 (from statement 2) as factors. Hence, C is the answer.
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Re: multiple of 15 [#permalink] New post 31 Jan 2011, 01:38
Q: Is n a multiple of 15?

This can be solved by prime factors.

Question can be rephrased as; are at least both 3 and 5 prime factors of n

3 and 5 are factors of 15. If n also contains at least both 3 and 5 as factors, it must be divided by 15.

1. n is a multiple of 20.
Prime factors of 20 are 2*2*5

This tells us that n is definitely a multiple of 5. But, there is no 3 among its factors. We must have at least both 3 and 5 as factors for n to be a definite multiple of 15.

We also can't definitely tell that n is not a multiple of 15.

e.g.
n=40- Multiple of 20. NOT a Multiple of 15.
n=60- Multiple of 20. Also a multiple of 15.

So, the prime factors 2*2*5 tell us that n is NOT definitely a multiple of 15. It may or may not be a multiple of 15. NOT SUFFICIENT.

2. (n+6) is a multiple of 3.

6 is a muliple of 3, so n must also be a multiple of 3.

if (a+b) is a muliple of x and b is a multiple of x, then "a" must be a multiple of x
if (a-b) is a muliple of x and b is a multiple of x, then "a" must be a multiple of x
conversely also true,
if "a" is a multiple of x and "b" is a multiple of x,
then (a+b) must be a multiple of x
also; (a-b) must be a mutiple of x

So, we know n is definitely a multiple of 3. i.e. 3 is a factor of n. But, n is not necessarily a multiple of 15.

For "n" to be a multiple of 15, it must have at least both 3 and 5 as factors.

e.g.

6- multiple of 3. Not a multiple of 15.
30- multiple of 3. Also a multiple of 15.
NOT SUFFICIENT.

Using both statements;
We know 5 and 3 are both factors of n. Thus, n must be a multiple of 15.

SUFFICIENT.

Ans: C
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Re: multiple of 15 [#permalink] New post 31 Jan 2011, 02:47
Expert's post
Wayxi wrote:
Is the integer n a mulltiple of 15 ?

1) n is a multiple of 20

2) n + 6 is a multiple of 3


Please explain how to derive the answer to this question. Thanks


Is the integer n a mulltiple of 15 ?

(1) n is a multiple of 20 --> now, if \(n=0\) then the answer will be YES but if \(n=20\) then the answer will be NO. Not sufficient.

But from this statement we can derive that as \(n\) is a multiple of 20 then it's a multiple of 5.

2) n + 6 is a multiple of 3 --> again, if \(n=0\) then the answer will be YES but if \(n=3\) then the answer will be NO. Not sufficient.

But from this statement we can derive that \(n\) is a multiple of 3 (\(n+6=3q\) --> \(n=3(q-2)\), for some integer \(q\):).

(1)+(2) \(n\) is a multiple of both 5 and 3 thus it must be a multiple of 3*5=15. Sufficient.

Answer: C.
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Re: multiple of 15 [#permalink] New post 31 Jan 2011, 07:04
Thanks a lot guys. Really put it into perspective. I was a little confused with how to use the second statement but how fluke broke it down makes sense. I remember the property that mgmat gave. When a is a multiple of x and b is a mutiple of x. a + b will be a multiple of x.
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Re: multiple of 15 [#permalink] New post 07 Feb 2011, 21:25
Bunuel
Dont mind me asking- but i have been noticing that you have a very different approach and i was wondering if this is something you picked up from a textbook or elsewhere. I would have approached it the way fluke approached the problem.
Re: multiple of 15   [#permalink] 07 Feb 2011, 21:25
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