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I've used just common sense here, since I'm not getting anywhere with traditional ways.

(A) St 1 is sufficient. Because, since 18 = 3*3*2, 5 and 18 have no factors in common.

Which means for 5n/18 to be an integer, n must be a multiple of 18!
Hence sufficient.

(B) is not sufficient because 3 and 18 have common factors, so n may be a multiple of 18, the only thing certain here is that it will be a multiple of 6. Therefore not sufficient.

(1) If 5n/18 is an integer, call it k, we know that n/18=k/5, which may or may not be an integer. NOT SUFF

If (5n/18)=k=>5n=18k=>n=(18k/5) since both n and k are INTEGERS then k is multiple of 5 and respectively n is a multiple of 18 and an INTEGER so A) is SUFF byitself,

lets see..the stem says nothing about N being an integer...soo

1) 5 N/18 - integer...well N can be an integer multiple of 18 or it can be 18M/5 where M is an integer not divisible by 5...Insuff

2) simplify this to N/6, so N has prime factors 2 and 3, we still dont know if it has enuff 3s or not...but we noe for sure that N is not a fraction...

combining them we now now that N is not a fraction and that its a mutliple of 18....

C it is...

Answer woud be A if we are told in the stem that N is an integer..

Re: Is n/18 an integer? (1) 5n/18 is an integer. (2) 3n/18 is an [#permalink]

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18 Nov 2013, 03:54

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5n/18=integer If we consider n=18M/5 Where M can be number which is not divisible by 5 .. If we consider this case, then 5(18M/5)/18=integer is not fulfilling sinceM is not integer , to fulfill this M must be integer hence N is integer..

From F.S 1, we know that \(\frac{5n}{18}\) is an integer. For\(\frac{n}{18} = 1\), we have a YES. Again, for \(\frac{n}{18} = \frac{1}{5}\) , we have a NO.Insufficient.

From F.S 2, we know that\(\frac{3n}{18}\) is an integer. For \(\frac{n}{18} = 1\), we have a YES,but for \(\frac{n}{18} = \frac{1}{3}\) , we have a NO.Insufficient.

Taking both together, we know that from F.S 1, either\(\frac{n}{18} = \frac{k}{5}\) or \(\frac{n}{18} = p\) , where k,p are integers and k and 5 are co-primes.

But, for \(\frac{n}{18} = \frac{k}{5}\), \(\frac{3*n}{18} = \frac{3*k}{5}\) and it will not be an integer. Thus, \(\frac{n}{18} = p\) can the only be form possible.

We need to find whether n is a multiple of 18 or n= 18I for some integer I

St 1: 5n/18= Integer or n = 18*Integer/ 5 Now if Integer = 2, then n =7.2 and n/18 is not an integer

but integer = 5 then n = 18 and 18/18 is an integer ------Possible values of n = 3.6,7.2,10.8,14.4,18......

So A and D ruled out

St 2: we see that n is a multiple of 6 so possible values of n =6,12,18...... If n= 6 then n/18 is not an integer but if n=18 then n/18 is an integer. So option B ruled out

Combining we get possible values of n =18,36,54 and so on

Hence Ans C
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From 1: 5n/18 = a (an integer) => n/18 = a/5 (we cannot be sure if this is an interger) From 2: 3n/18 = b (an integer) => n/18 = b/3 (we cannot be sure if this is an interger) Combining 1&2: 2*(statement 2) - (statement 1) => 6n/18 - 5n/18 = 18*2*b - 18*a => n/18 = 18(2b-a) {this we know for sure is an integer} Thus, Option C

It's C, but I think this is as clear as it can be explained:

S1: \(\frac{5n}{18}=K1\) (Letting K1 be an integer)

Then, isolating n we get \(n=\frac{18K1}{5}\)

Going back to the original question if n is divisible by 18, based on this answer is "Maybe" (provided that K1 is a multiple of 5)

Insufficient.

S2: \(\frac{3n}{18}=K2\) (Letting K2 be an integer)

Then, isolating n we get \(n=\frac{18K2}{3}\), reducing \(n=6K2\)

Going back to the original question if n is divisible by 18, based on this answer is "Maybe" (Provided that K2 is a multiple of 3)

Insufficient.

S1 & S2: We will make the expressions for n equal:

\(\frac{18K1}{5}=6K2\), simplifying \(K2=\frac{3K1}{5}\) The key here is understanding that K1 and K2 MUST be integers. As such, the "Maybes" of S1 and S2 are proven to be true. From the expression above, K2 is a multiple of 3 and K1 is a multiple of 5.

This is perhaps an extended explanation, but I think is clear enough.

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