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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.

Umm...I am having 1 query ...very simple

I don't want to get into such complex solution my only query is ...

S 1 : tells 5n/18 is an integer - 5 is prime, hence it must be clear that "n" will be divisible by 18 , in that case n/18 will be an integer .

then why some folks are not mentioning A wrong ??

Is i am missing any logic plz explain ?
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I don't want to get into such complex solution my only query is ...

S 1 : tells 5n/18 is an integer - 5 is prime, hence it must be clear that "n" will be divisible by 18 , in that case n/18 will be an integer .

then why some folks are not mentioning A wrong ??

Is i am missing any logic plz explain ?

You are missing that we are NOT told that n is an integer. So, 5n/18=integer, does not necessarily mean that n is a multiple of 18, if for example n=18/5, then it's not so.

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.

I don't want to get into such complex solution my only query is ...

S 1 : tells 5n/18 is an integer - 5 is prime, hence it must be clear that "n" will be divisible by 18 , in that case n/18 will be an integer .

then why some folks are not mentioning A wrong ??

Is i am missing any logic plz explain ?

You are missing that we are NOT told that n is an integer. So, 5n/18=integer, does not necessarily mean that n is a multiple of 18, if for example n=18/5, then it's not so.

Hope it's clear.

Than you very much for giving me an insight...i will take care this in future.
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This was my reasoning as well but I don't get how 'C' can be the answer. I mean for 5n to be divisible by 18 it must have the same prime factors as 18. In which case shouldn't 'n' have the same prime factors as 18 thereby making n/18 an integer?

mikki0000 wrote:

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.

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.

Answer is C

I can see how these statements alone are insufficient.

Just testing numbers gives that. For 5n/18 to be an integer 5n has to be a multiple of 18, but for instance if 5n equals 18 then n equals 3, something so that wont be divisible by 18. Same goes for the second statement.

But i cant find any clear explanation that i understand why these 2 taken together are sufficient. Can someone use examples please? thank you in advance.

Bunuel does an excellent job explaining it earlier in the thread, but I'll give it a shot (I find it helps my own understanding).

In statement 1, we are told that 5n/18 = Int. n = (18/5) * Integer. We do not know if this results in n being an int or not. Insufficient. In statement 2, we are told that 3n/18 = Int. n = 6 * integer. In this statement, we know that n is an integer (int * int = int). However, that doesn't mean that n/18 is an integer (question stem). This can be illustrated by picking n = 6 and n = 18; one yields 1/3 (not int), and one yields 1 (integer).

When we combine 1 + 2, it is sufficient: From statement 2, we know that n MUST BE an integer (recall n = 6 * int, which must yield an int). Now by looking at statement 1, we notice that 5n/18 = integer. If n is an integer, it MUST BE a multiple of 18, because 5 itself does not share any prime factors with 18. Therefore, the only way that 5n/18 can be an integer with n being an integer is for n to be a multiple of 18.

If the qustion says division result (n/18) is an integer (no fractional part) , shoul we consider decimal values also?

Hi SoumiyaGoutham,

Integers by definition are Non-Fractional and Non-Decimal type Numbers. So (5n/18) should be an Integer doesn't ascertain the value of n which theoretically can be a fraction of two integers (or Decimal value) as well e.g. for n= 18/5, (5n/18) will be an Integer

"An integer (from the Latin integer meaning "whole")[note 1] is a number that can be written without a fractional component. For example, 21, 4, 0, and −2048 are integers, while 9.75, 5½, and √2 are not.

The set of integers consists of zero (0), the natural numbers (1, 2, 3, …), also called whole numbers or counting numbers,[1] and their additive inverses (the negative integers, i.e. −1, −2, −3, …). This is often denoted by a boldface Z ("Z") or blackboard bold \mathbb{Z} (Unicode U+2124 ℤ) standing for the German word Zahlen ([ˈtsaːlən], "numbers").[2][3] ℤ is a subset of the sets of rational and real numbers and, like the natural numbers, is countably infinite.

The integers form the smallest group and the smallest ring containing the natural numbers. In algebraic number theory, the integers are sometimes called rational integers to distinguish them from the more general algebraic integers. In fact, the (rational) integers are the algebraic integers that are also rational numbers."

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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

Thanks for the solution, really simple. But I tried to make your solution more understandble.

Let n/18=p, (we don't know if p is integer), 1) 5p=a, a is integer, but p=a/5 not necessarily interger 2) 3p=b, b is integer, but p=b/3 not necessarily interger equation 2*2-equation 1: on the left: 2*3p-5p=p on the right: 2b-a so we have the equition p=2b-a, which is clearly integer.

Condition 1: n is a multiple of 18 like 36, e.g.: (5 x 36)/18 ==> n/18 = 36/18 ==> is an integer.

Condition 2: n is a fraction with the numerator a multiple of 18 and denominator 5 like 36/5, e.g.: (5x(36/5))/18 ==> n/18 = (36/5)/18 ==> not an integer.

Condition 3: n =0.

INSUFF.

St 2: 3n/18 is an integer. True only if

Condition 1: n is a multiple of 18 like 18, e.g.: (3 x 18)/18 ==> n/18 = 18/18 ==> is an integer.

Condition 2: n is a fraction with the numerator a multiple of 18 and denominator is 3 like 18/3, e.g.: (3 x (18/3))/18 ==> n/18 = (18/3)/18 ==> not an integer.

Condition 3: n = 0.

INSUFF.

Combined: Since Conditions 2 of statements 1 and 2 are mutually exclusive(denominator of n can't be 3 and 5 simultaneously), either condition 1 or condition 3 is true, in either case, n/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 \(\frac{n}{18}=\frac{1}{5}\neq{integer}\).

Two different answers. Not sufficient.

(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 \(\frac{n}{18}=\frac{1}{3}\neq{integer}\).

Two different answers. Not sufficient.

(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.

Answer: C.

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

Hope it helps.

Hi Bunuel,

Just to get more clarity, what would be your answer if the second statement is changed to- 7n/18 is an integer?

Actually i had a different approach to the answer:

Take n/18 as a constant R

Now, since 5R is an integer, the only way R is not an integer is when R itself is a simplified fraction with only 5 as denominator, which is a possibility. Similarly for 3R, a fraction with only 3 as denominator

Using both, i know it is not possible that both 5R and 3R are integers when R is not, because my above cases can only be true by having only one digit (5 or 3) as denominator. Thus, R has to be an integer itself.

5n/18 = K (K considered as some Integer). Therefore n = 18k/5. Hence if k = 1, then n = 18/5 i.e. 3.6. Hence n/18 is not an an integer. Hence insufficient. But if we consider k as 5, the n = 18 x 5/18 = 18. Then n/18 = 18/18 = 1.

3n/18 = K (K considered as some Integer). Therefore n = 6K. Therefore if K = 1 or 2, Then 6 x 1/18 or 6 x 2/18 is not an integer. Hence insufficient., But if we consider k as 3, then n = 6 x3 = 18 and hence n = 18/18 = 1.

Considering both the stamens we know that n = 18, 36, 54 etc Hence "C"

I think it can be explained in an even more straightforward fashion-- tell me if i'm wrong.

is n/18 an integer is equivalent to asking is x an integer.

1. 5x is an integer. For this to be case, x can either be an integer, or be equal to 1/5-- if it were to be a non integer. X can here take integer, or non integer characteristics. Not enough.

2. 3x is an integer. For this to be case, x can either be an integer, or be equal to 1/3-- if it were to be a non integer. X can here take integer, or non integer characteristics. Not enough.

1&2: The two given information would contradict another if x were to be a non integer. if X were to be 1/5, the 3x would not be an integer, and vice versa.

Statement 1: 5n is a multiple of 18 or n is a multiple of 3.6. Not sufficient Statement 2: 3n is a multiple of 18 or n is a multiple of 6. Not sufficient

Both 1 and 2. n is a multiple of both 3.6 and 6 i.e, n is a multiple of 18. Sufficient

We want to know whether \(\frac{n}{18}=Integer =>\) whether \(n=18I\) (for some integer \(I\)).

This implies two things A) whether \(n\) is integer? & B) if \(n\) is an integer then whether it is a multiple of \(18\)?

Statement 1: \(\frac{5n}{18} = Integer => n=\frac{18k}{5}\) (for some integer \(k\)), if \(k\) is a multiple of \(5\), then \(n\) is an integer and a multiple of \(18\) and if \(k\) is not a multiple of \(5\), then it is not. Insufficient

Statement 2: \(\frac{3n}{18}=Integer => n=6q\) (for some integer \(q\)). This CONFIRMS that \(n\) is an INTEGER. but we do not know yet that \(n\) is a multiple of \(18\). Insufficient

Combining 1 & 2: because \(n\) has to be an integer so we have, \(n=\frac{18k}{5}=Integer\). This is only possible if \(k\) is a multiple of \(5\). Hence \(n\) will be a multiple of \(18\). Sufficient

The video Solution with DS handling technique and commonly made mistakes

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