one more useful thread from the past

what would be your take on here?
i describe mine --> n/15?
st(1) n/(4*5) Not Sufficient as we know that n can be divided by 5 BUT we are looking for the denominator 3*5
st(2) (n+6)/3 Not Sufficient as we can only simplify to 3(n/3 +2)/3 is an integer and (n/3 + 2) must be integer; n is divided by 3 BUT we don't know about its divisibility by 5;
Combined st(1&2): n/5 and n/3 both integers so n/15 is also an integer Sufficient

IOM C

dferm wrote:Is the integer n a multiple of 15?

(1) n is a multilple of 20
(2) n+6 is a multiple of 3

Please assist...

Thanks...

An interesting way to approach this problem is to re-phrase the prompt in terms of prime factorization.

"Is the integer n a multiple of 15?" is really asking "Does n have both 3 and 5 as factors?" or alternately "Is n a multiple of both 3 and 5?"

Statement (1) tells you that n is a multiple of 20. You want to know if n has 3 and 5 as factors. Well, the 5 is taken care of, because 5 is a factor of 20. But what about the 3? It's not clear.

You can also illustrate this by picking numbers that fit the condition of St (1). Examples would be n = 20, 40, 60, 80, etc
All of those have 5 as a factor, but not all of them have 3 as a factor. INSUFFICIENT.

Statement (2) says n+6 is a multiple of 3. The tricky thing here is to realize that 6 is a multiple of 3, and thus if n+6 is a multiple of 3, n itself must also be a multiple of 3.

Again, you can test numbers to verify this. n could equal 0, 3, 6, 9, etc. All those values of n are already multiples of 3.

So St (2) really just says "n is a multiple of 3."

Unfortunately, we don't know about the 5, so we can't say if n is a multiple of 15. INSUFFICIENT.

Together, (1) tells us n has 5 as a factor and (2) tells us n has 3 as a factor. Therefore, since n has both 5 and 3 as factors, it must also have 3*5 = 15 as a factor. SUFFICIENT

Ans: C

On questions that ask about multiples and factors in general, prime factorization is a great shortcut.
Here's a post I wrote that goes into more detail (and involves questions similar to this one):
https://www.knewton.com/blog/gmat/quant- ... n-the-gmat

Enjoy! Hope that helps.

C it is.

Obviously niether is suff on its own.

Using 1, n has one factor 5, for it to be div by 15=5*3 we need it to be a factor of 3.

Using 2: n+6 = 3k => n = 3(k-2) so we knew n is a multiple of 3.

Combining workd

dferm wrote:Is the integer n a multiple of 15?

(1) n is a multilple of 20
(2) n+6 is a multiple of 3

Please assist...

Thanks...