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For \(\frac{n}{18}\) to be an integer, n must be a multiple of 18. \(18 = 2*3^2\)

1) As 5 is not one of the factors of 18, n must be a multiple of 18 for \(\frac{5n}{18}\) to be an integer. Therefore, sufficient.

2) Because 3 is one of the factors, you cannot know whether n/18 is an integer. If n=6, then the condition would be satisfied and the answer would be no (Is n/18 an integer?). If n=18, the condition would also be satisfied and the answer would be yes. Insufficient.

Kool trap qn there.... This one actually makes think whether to go back to strategically solving every DS in eqns, rather than making assumptions and go wrong easily...

For \(\frac{n}{18}\) to be an integer, n must be a multiple of 18. \(18 = 2*3^2\)

1) As 5 is not one of the factors of 18, n must be a multiple of 18 for \(\frac{5n}{18}\) to be an integer. Therefore, sufficient.

2) Because 3 is one of the factors, you cannot know whether n/18 is an integer. If n=6, then the condition would be satisfied and the answer would be no (Is n/18 an integer?). If n=18, the condition would also be satisfied and the answer would be yes. Insufficient.

Therefore, A.

I also chose A. Can someone explain why A is insufficient?

For \(\frac{n}{18}\) to be an integer, n must be a multiple of 18. \(18 = 2*3^2\)

1) As 5 is not one of the factors of 18, n must be a multiple of 18 for \(\frac{5n}{18}\) to be an integer. Therefore, sufficient.

2) Because 3 is one of the factors, you cannot know whether n/18 is an integer. If n=6, then the condition would be satisfied and the answer would be no (Is n/18 an integer?). If n=18, the condition would also be satisfied and the answer would be yes. Insufficient.

Therefore, A.

I also chose A. Can someone explain why A is insufficient?

In 1) n could be any multiple of 3.6 or 18.
_________________

In statement 1, n can be 3.6. - A is insufficient; In statement 2, n can be 6. - B is insufficient;

When we consider both statements together, n could be any number divisible by 18. (18, 36, 54, 72 ... etc). I couldn't find a number n that could could meet both statements together, but is not divisible by 18. So 'C' is sufficient.

In statement 1, n can be 3.6. - A is insufficient; In statement 2, n can be 6. - B is insufficient;

When we consider both statements together, n could be any number divisible by 18. (18, 36, 54, 72 ... etc). I couldn't find a number n that could could meet both statements together, but is not divisible by 18. So 'C' is sufficient.

in statement 1 i think 'n' can only be a multiple of 18 and not 3 or 6 can you pls explain how in statement 1 'n' can be a multiple of 3 or 6 and satisfy the condition that 5n/18 is an integer

In statement 1, n can be 3.6. - A is insufficient; In statement 2, n can be 6. - B is insufficient;

When we consider both statements together, n could be any number divisible by 18. (18, 36, 54, 72 ... etc). I couldn't find a number n that could could meet both statements together, but is not divisible by 18. So 'C' is sufficient.

in statement 1 i think 'n' can only be a multiple of 18 and not 3 or 6 can you pls explain how in statement 1 'n' can be a multiple of 3 or 6 and satisfy the condition that 5n/18 is an integer

I guess what he meant was not that n could be 3 or 6, but that n could be the number 3.6 or its multiples.

By the way, nice catch x2suresh. My factoring method blinded me from the possibility of n as a non integer.
_________________

In statement 1, n can be 3.6. - A is insufficient; In statement 2, n can be 6. - B is insufficient;

When we consider both statements together, n could be any number divisible by 18. (18, 36, 54, 72 ... etc). I couldn't find a number n that could could meet both statements together, but is not divisible by 18. So 'C' is sufficient.

in statement 1 i think 'n' can only be a multiple of 18 and not 3 or 6 can you pls explain how in statement 1 'n' can be a multiple of 3 or 6 and satisfy the condition that 5n/18 is an integer

5n/18 = i (integer) n/18 = i/5 --(1) when i=1 its not integer when i=5 its integer not sufficient

II. 3n/18 is an integer II. 3n/18 is an integer 5n/18 = j (integer) n/18 = j/6 --> (2) when j=1 its not integer when j=6 its integer

from (1) and (2)

i/5 =j/6

6i= 5 j from the above equation it is clear that i must be multiple of 5 and j must be multiple of 6

so i= 5k (k-integer)

n/18 = i/5 = 5k/5= k=integer

C

Nice proof. Most of the time, when the question looks simple, it makes us to think in a wring direction easily. In this case, every where in the question and stmts, it is referring to Integers, we forget the fact that n could be factors as well.