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mgmat digits 700 question

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Source: — Data Sufficiency |
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by codesnooker » Wed Apr 09, 2008 11:44 pm
Are you sure that statement 1: 5n/18?

If according to book the answer is C, then here numerator should be factor of 18.

HOW?

Let me try to explain...

To prove n/18 is an integer, we must need to prove that n is the multiple of 18.

From the current statement 1: 5n/18 is an integer. So if 5n/18 to be an integer, then n must be the multiple of 18, otherwise it can't be an integer.
Hence, statement 1 is alone to satisfy the that n/18 is an integer.

From statement 2: 3n/18 is an integer.

Here if 3n/18 is an integer than n should be multiple of 6. So now if n is multiple of 6, it means it could be 6, 12, 18 etc. Its not sufficient if n would be equal to 6.

So statement 2nd is not alone sufficient.

According to me, if you have written the statement correctly then answer should be A. Not C.

Let any Manhattan instructor or any other instructor reply about this.
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by codesnooker » Wed Apr 09, 2008 11:45 pm
Can you post the explanation given in the book?
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by simplyjat » Thu Apr 10, 2008 12:01 am
I did the same mistake. Actually the question never tell that n is an integer.
So if 5n/18 is an integer, it does not mean that n is an integer. Take n=18/5 and 5n/18 become 1, and clearly 18/5 in not an integer...
simplyjat
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by codesnooker » Thu Apr 10, 2008 4:48 am
Hmm. Indeed you are correct. My bad!!! How could I missed that, in some earlier post to engin, I already asked him to consider first 1/multiple of the given number as the factor of the given number. And now I missed myself . :shock: lol

Okay then lets try to solve it again.

Lets say n = 18/5 and its not an integer. Then n/18 would not be an integer.
Hence, statement 1 is alone not sufficient.

For statement 2, we have already proved that in my earlier posts.

Lets try to take both statements together.

If n=18/5, then 5n/18 is an integer. OK
but putting n=18/5 in 3n/18 will contradict the 2nd statement. Hence n can't be equal to 18/5.

Statement 1, also reveals that n could be 18x/5 where x = 1, 2, 3, ....

and

Similarly Statement 2, reveals that n could be 18y/3 where y = 1, 2, 3....

Now we need to write a series of 18x/5 and 18y/3 and need to find 1st value where it satisfy both the statements.

The first value would be 270/15. (You can check that, as the denominator of both statements are different, so you need to make equal by multiplying and dividing 1st series by 3 and second series by 5)

The 2nd value would be 540/15...


So the common series would be 270/15, 540/18, 810/15.....

When we simplify it, then it would reduced to 18, 36, 54..... which are multiple of 18.

So, that is the thing that we need to prove that n is the multiple of 18.

Hence, n/18 is an integer, proved by both statements.

Now Enjoy
:D
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by codesnooker » Thu Apr 10, 2008 5:00 am
Engin, one request to you. Please use spoiler to hide the answer. It would help everyone to prepare better.

Check its usages:-

https://www.beatthegmat.com/new-spoilers-t5302.html
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