Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink]

Show Tags

23 Jul 2012, 12:48

1

This post was BOOKMARKED

Great question. Made me think for a while.

My answer is D.) "K is a multiple of 9".

We know that M is always divisible by 18 which means that M is always divisible by 9. This implies that the sum of the digits of M (also referred to as k) will ALWAYs be divisible by 9 (refer to the divisibility rules if you want confirmation).

I kept thinking 0 was a loophole until I realized that 0 is a multiple of ALL integers so in the case k=0 (occurs when n=0), k is still a multiple of 9. I also got stuck for a bit on answer choice E.) "K is a multiple of 6" until some plug-n-chug at n=1 disproved this answer.

I'd be very interested in seeing how other people solved this - please post if you used a different route of thinking.

Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink]

Show Tags

14 Aug 2012, 23:10

Club909 wrote:

Great question. Made me think for a while.

My answer is D.) "K is a multiple of 9".

We know that M is always divisible by 18 which means that M is always divisible by 9. This implies that the sum of the digits of M (also referred to as k) will ALWAYs be divisible by 9 (refer to the divisibility rules if you want confirmation).

I kept thinking 0 was a loophole until I realized that 0 is a multiple of ALL integers so in the case k=0 (occurs when n=0), k is still a multiple of 9. I also got stuck for a bit on answer choice E.) "K is a multiple of 6" until some plug-n-chug at n=1 disproved this answer.

I'd be very interested in seeing how other people solved this - please post if you used a different route of thinking.

D says that k has to be a multiple of 9 K = sum of the digits of M

so lets a couple of cases

we know m = 18 n

when n=0 m=0 so k =0 and k is a multiple of 9 , D is true

when n=1,2...6 m= 18, 36,....108 so k = 9 again K is a multiple of 9 , D is again true

when n = -1 or -2 or -6 then m = -18 or -36 or -108 then k = 7 or 3 or 7 ..but now K is not a multiple of 9 ??

so how can D always be true ??

Please note question does not mention that n is a positive integer or M is a positive integer . if n is a negative integer as shown above then m will be negative and the sum of the digits of M will not always be 9 so please do explain how D is always true ?? _________________

We know that M is always divisible by 18 which means that M is always divisible by 9. This implies that the sum of the digits of M (also referred to as k) will ALWAYs be divisible by 9 (refer to the divisibility rules if you want confirmation).

I kept thinking 0 was a loophole until I realized that 0 is a multiple of ALL integers so in the case k=0 (occurs when n=0), k is still a multiple of 9. I also got stuck for a bit on answer choice E.) "K is a multiple of 6" until some plug-n-chug at n=1 disproved this answer.

I'd be very interested in seeing how other people solved this - please post if you used a different route of thinking.

D says that k has to be a multiple of 9 K = sum of the digits of M

so lets a couple of cases

we know m = 18 n

when n=0 m=0 so k =0 and k is a multiple of 9 , D is true

when n=1,2...6 m= 18, 36,....108 so k = 9 again K is a multiple of 9 , D is again true

when n = -1 or -2 or -6 then m = -18 or -36 or -108 then k = 7 or 3 or 7 ..but now K is not a multiple of 9 ??

so how can D always be true ??

Please note question does not mention that n is a positive integer or M is a positive integer . if n is a negative integer as shown above then m will be negative and the sum of the digits of M will not always be 9 so please do explain how D is always true ??

The sum of the digits of -18 is still 9 (1+8) not not 7 (-1+8).

Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink]

Show Tags

14 Aug 2012, 23:34

Bunuel wrote:

stne wrote:

Club909 wrote:

Great question. Made me think for a while.

My answer is D.) "K is a multiple of 9".

We know that M is always divisible by 18 which means that M is always divisible by 9. This implies that the sum of the digits of M (also referred to as k) will ALWAYs be divisible by 9 (refer to the divisibility rules if you want confirmation).

I kept thinking 0 was a loophole until I realized that 0 is a multiple of ALL integers so in the case k=0 (occurs when n=0), k is still a multiple of 9. I also got stuck for a bit on answer choice E.) "K is a multiple of 6" until some plug-n-chug at n=1 disproved this answer.

I'd be very interested in seeing how other people solved this - please post if you used a different route of thinking.

D says that k has to be a multiple of 9 K = sum of the digits of M

so lets a couple of cases

we know m = 18 n

when n=0 m=0 so k =0 and k is a multiple of 9 , D is true

when n=1,2...6 m= 18, 36,....108 so k = 9 again K is a multiple of 9 , D is again true

when n = -1 or -2 or -6 then m = -18 or -36 or -108 then k = 7 or 3 or 7 ..but now K is not a multiple of 9 ??

so how can D always be true ??

Please note question does not mention that n is a positive integer or M is a positive integer . if n is a negative integer as shown above then m will be negative and the sum of the digits of M will not always be 9 so please do explain how D is always true ??

The sum of the digits of -18 is still 9 (1+8) not not 7 (-1+8).

Hope it's clear.

Ok, if - 18 = 1+8 then D is always true , Got it _________________

Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink]

Show Tags

08 Dec 2013, 01:57

2

This post received KUDOS

lajulajay wrote:

Can someone post an example of a case when C is not true?

Hello lajulajay

Let try n = 11 ==> m = 18*11 = 198 ==> k = 1 + 9 + 8 = 18 ==> 2k = 36

But 198 / 36 = 5.5 ==> C is not always correct.

Hope it helps. _________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink]

Show Tags

19 Jun 2015, 08:42

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

Time is a weird concept. It can stretch for seemingly forever (like when you are watching the “Time to destination” clock mid-flight) and it can compress and...