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Manager  Joined: 28 Jan 2006
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Difficulty:   95% (hard)

Question Stats: 23% (01:55) correct 77% (01:23) wrong based on 1381 sessions

HideShow timer Statistics Is n/18 an integer?

(1) 5n/18 is an integer.

(2) 3n/18 is an integer.

SOLUTION IS HERE: https://gmatclub.com/forum/is-n-18-an-i ... l#p1341451

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shinewine
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Re: Is n/18 an integer?  [#permalink]

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bparrish89 wrote:
mau5 wrote:
shinewine wrote:
Is n/18 an integer?

(1) 5n/18 is an integer.

(2) 3n/18 is an 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.

C.

I am absolutely stumped on understanding the above explanations. My thought process was to break down the denominator into its primes --> 3^2 and 2, then identify whether N has those same characteristics, leading my to choose A.

I know this is a vague response, but any chance you could help me understand how A is wrong and how the 2 combined MUST form an integer.

Is n/18 an integer?

Notice that we are NOT told that n is an integer.

(1) 5n/18 is an integer:

If $$\frac{5n}{18}=0$$, then $$n=0$$ and $$\frac{n}{18}=0=integer$$;
If $$\frac{5n}{18}=1$$, then $$n=\frac{18}{5}$$ and $$\frac{n}{18}=\frac{1}{5}\neq{integer}$$.

(2) 3n/18 is an integer --> $$\frac{3n}{18}=\frac{n}{6}=integer$$ --> $$n=6*integer=integer$$. So, this statement implies that n is a multiple of 6.

If $$n=0$$, then $$\frac{n}{18}=0=integer$$
If $$n=6$$, then $$\frac{n}{18}=\frac{1}{3}\neq{integer}$$.

(1)+(2) Since from (2) we have that n is an integer, then from (1) it follows that it must be a multiple of 18. Sufficient.

If we were told in the stem that n is an integer, then the answer would be A.

Hope it helps.
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Q: n = 18 x k where k = integer.

S1: n = 18xm/5 where m = integer.

Therefore, n/18 = m/5 Not sufficient.

S2: n= 18 x s/ 3 where s = integer.

=> n/18 = s/3; Not sufficient.

S1 & S2:
n/18 = m/5
n/18 = s/3

or m/5 = s/3
or 3m = 5s

Which means m & s are multiples of both 5 and 3.

Therefore n/18 is an integer.

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VP  Joined: 02 Jun 2006
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Re: Is n/18 an integer?  [#permalink]

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If n/18= 0 and n/18 = 1/5,

In one case, we get
5n/18 = 0 integer. n/18 => integer...

5n/18 =1 integer, n/18 => not integer...

I don't understand how its A?

X & Y wrote:
Getting A

St 1: Suff
St 2: Insuff
Manager  Joined: 09 Aug 2005
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Re: Is n/18 an integer?  [#permalink]

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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.

MG
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Re: Is n/18 an integer?  [#permalink]

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We don't know that n is an integer!
VP  Joined: 02 Jun 2006
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Kevin,
Can you elaborate a little bit? I think I am losing my mind....can't seem to figure this out for some reason? Are you getting A too?

kevincan wrote:
We don't know that n is an integer!
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Ah so then I'm wrong....

i.e. N can be 18/5 to begin with... thats true.

But Haas, going back to your solution, if N is a multiple of both 5 and 3, how can you be sure it is a multiple of 18??

It only proves it can be a multiple of 15. It still may or may not be a multiple of 18.

Does that mean hte answer is E?

MG
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Re: Is n/18 an integer?  [#permalink]

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shinewine wrote:
Is n/18 an integer?

(1) 5n/18 is an integer.

(2) 3n/18 is an integer.

n/18 can be x/5 in 1 and x/3 in 2 where x is some number

taking both into consideration we get nothing

E it is
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Re: Is n/18 an integer?  [#permalink]

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I believe it is A...

I think it was said before

stmnt1- 5 shares no common factors with 18. This means that N must be a direct factor of 18. Suff

stmnt2- Insuff for reasons explained above.
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(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

(2) Similarly, if 3n/18 is an integer, call it m, we know that n/18= m/3, which may or may not be an integer NOT SUFF

(1) and (2): For some integers k and m, k/5=m/3 => 3k=5m =>k=5m/3.

Thus m is a multiple of 3, so n/18=m/3 is an integer SUFF
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Re: Is n/18 an integer?  [#permalink]

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(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,
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Re: Is n/18 an integer?  [#permalink]

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rdw28 wrote:
Kevincan,

There is no way possible that 5n/18 could be an integer if n were not an integer.

There is: consider n/18 = 1/5. Then 5 (n/18) = 5. (1/5) = 1.
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Re: Is n/18 an integer?  [#permalink]

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shinewine wrote:
Is n/18 an integer?

(1) 5n/18 is an integer.

(2) 3n/18 is an integer.

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..
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Re: Is n/18 an integer?  [#permalink]

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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..

Thus A should be OA.
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shinewine wrote:
Is n/18 an integer?

(1) 5n/18 is an integer.

(2) 3n/18 is an 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.

C.
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shinewine wrote:
Is n/18 an integer?

(1) 5n/18 is an integer.

(2) 3n/18 is an integer.

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
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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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shinewine wrote:
Is n/18 an integer?

(1) 5n/18 is an integer.

(2) 3n/18 is an integer.

Statement I is insufficient:

n = 36 (YES)
n = 18/5 (NO)

Statement II is insufficient:

n = 36 (YES)
n = 6 (NO)

Combining is sufficient:

Since 5 and 3 are co primes to each other n will have to be a multiple of 18 for 5n/18 and 3n/18 to be an integer.

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