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yes. 5n/18 is integer=>n =18 or its multiple and hence, n/18 is integer. 3n/18 is integer=>n=6 or its multiple=>sometimes integer and sometimes not. Hence insufficient.

yes. 5n/18 is integer=>n =18 or its multiple and hence, n/18 is integer. 3n/18 is integer=>n=6 or its multiple=>sometimes integer and sometimes not. Hence insufficient.

Answer A.

Stmt 1 would be suff if n were integer. n can be, say, 18/5.

ok...the answer is C. see below for explanation. I get part 1 and 2 counterexamples, but Im confused with combining 1 and 2 and LCM stuff (bold part). Any effort in putting those in simple words would be much appreciated.

------------------------------------------ (1) INSUFFICIENT: We are told that 5n/18 is an integer. This does not allow us to determine whether n/18 is an integer. We can come up with one example where 5n/18 is an integer and where n/18 is NOT an integer. We can come up with another example where 5n/18 is an integer and where n/18 IS an integer.

Let's first look at an example where 5n/18 is equal to the integer 1.

If 5n/18 = 1, then n/18 =1/5. In this case n/18 is NOT an integer.

Let's next look at an example where 5n/18 is equal to the integer 15.

If 5n/18 = 15, then n/18= 3. In this case n/18 IS an integer.

Thus, Statement (1) is NOT sufficient. (2) INSUFFICIENT: We can use the same reasoning for Statement (2) that we did for statement (1). If 3n/18 is equal to the integer 1, then n/18 is NOT an integer. If 3n/18 is equal to the integer 9, then n/18 IS an integer.

(1) AND (2) SUFFICIENT: If 5n/18 and 3n/18 are both integers, n/18 must itself be an integer. Let's test some examples to see why this is the case.

The first possible value of n is 18, since this is the first value of n that ensures that both 5n/18 and 3n/18 are integers. If n = 18, then n/18 is an integer. Another possible value of n is 36. (This value also ensures that both 5n/18 and 3n/18 are integers). If n = 36, then n/18 is an integer.

A pattern begins to emerge: the fact that 5n/18 AND 3n/18 are both integers limits the possible values of n to multiples of 18. Since n must be a multiple of 18, we know that n/18 must be an integer. The correct answer is C.

Another way to understand this solution is to note that according to (1), n = (18/5)*integer, and according to (2), n = 6*integer. In other words, n is a multiple of both 18/5 and 6. The least common multiple of these two numbers is 18. In order to see this, write 6 = 30/5. The LCM of the numerators 18 and 30 is 90. Therefore, the LCM of the fractions is 90/5 = 18.

Statement 1: 5n/18 is integer now we can also write this as 5 * n/18 as integer. Now we have not doubt that '5' is integer so for any thing 5x to be integer x has to be integer. here x is n/18 so since 5n/18 is integer n/18 is also an integer. - Sufficient.

Statement 2: 3n/18. Now in this case too we can apply the same analogy but here 18 is divisible by 3 (which is not possible in statement 1 as 18 is not divisible by 5) which means 3n/18 reduces to n/6. Now we can not certainly tell that if n/6 is integer then 3n/18 will also be integer. Hence this statement is not enough.