Is the integer n a multiple of 15?

yes right, that's typo error. I have updated my post.

I have one doubt for this question -Please help me to understand

1.n is a multiple of 20 ...I understand its not Sufficient

2.n+6 is a multiple of 3

considering n a multiple of 15 ,all possible multiples of 15 and +6 is always divisible by 3 ..So it should be sufficient ?
Not sure why OA- C ?

It should be the other way around: any multiple of 15 plus 6 is a multiple of 3, but it's possible \(n+6\) to be a multiple of 3 so that \(n\) not to be a multiple of 15. Consider \(n=3\).

Is the integer n a multiple of 15?

(1) n is a multiple of 20. If \(n=20\), then the answer is NO but if \(n=60\), then the answer is YES. Not sufficient.
From this statement though notice that \(n\) must be a multiple of 5.

(2) n+6 is a multiple of 3. If \(n=3\), then the answer is NO but if \(n=15\), then the answer is YES. Not sufficient.
From this statement though notice that \(n\) must be a multiple of 3, since \(n+6=3q\) --> \(n=3(q-2)\).

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

Answer: C.

Hope it's clear.

1 statement tels us that there are at least 2*2*5 as prime factors in n, but we are not sure that 3*5 are among the prime factors - so insufficient.
2 statement indicates that n is a multiple of 3 so it could be 0, 3, 15 ... - not sufficient
1+2 statements, here we see that n is a number which has 2*2*5 and 3 in its primes, so it must be a multiple of 15!

For a n to be multiple of 15, it has to be divisible its prime factors 3 and 5.

Statement 1 - n is multiple of 20 - divisible by prime factors 2 and 5. Not know about 3. Hence not sufficient.
Statement 2 - n+6 multiple of 3 - meaning n is divisible by 3. Not know about 5. Hence not sufficient.

Combing both. n is divisible by 2, 3, 5. Hence n will be multiple of 15.

The question asks whether n is a multiple of 15.

1) n is a multiple of 20. Clearly insufficient. However, notice that this means that 5,2,2 are prime factors of n. Thus, for n to be a multiple of 15, it also has to be a multiple of 3, to multiply with the 5 to get 15. We're looking to see if n is a multiple of 3.

2) n+6 is a multiple of 3. Notice this only says that n is a multiple of 3; if n+6 is a multiple of 3, then n+3 and n are also multiples of 3. On its own, it's insufficient, but it's precisely the information we were looking for from number 1.

Together, we have the information we need. Answer: C

"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

"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

Does N is multiple of 3*5 = 15 ?

From 1st statement:
\( \frac{n}{20}\) =\( \frac{n}{4*5}\)
Hence N must be multiple of 5, but we need additional factor 3 as well.

From 2nd statement:
\( \frac{n+6}{3}\) =\( \frac{n}{3}\) +\( \frac{6}{3}\)
Thus N is multiple of 3 yet we need additional factor 5 as well.

1st + 2nd statements give us necessitated factors 3 & 5.

Bunuel - I'm confused as to why the 2nd prompt is guaranteed to give us a factor of 3.

Given: n+6 = 3 * some integer

What if n + 6 = 6? Wouldn't that give us n=0 and thus we're missing the factor of 3 we need to answer the question?

ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. Zero is neither positive nor negative (the only one of this kind).

4. Zero is divisible by EVERY integer except 0 itself (\(\frac{x}{0} = 0\), so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer (\(x*0 = 0\), so 0 is a multiple of any number, x).

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 (\(x^0 = 1\))

9. \(0^0\) case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0), \(0^n = 0\).

11. If the exponent n is negative (n < 0), \(0^n\) is undefined, because \(0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}\), which is undefined. You CANNOT take 0 to the negative power.

12. \(0! = 1! = 1\).

I completely overlooked that it would still work anyway. Thank you for the help! Seems so obvious now.