GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Oct 2019, 01:49

### GMAT Club Daily Prep

#### 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

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

# If k is the sum of the digits of integer m, and m=18n, where

Author Message
TAGS:

### Hide Tags

Intern
Joined: 20 Jun 2011
Posts: 43
If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

Updated on: 14 Aug 2012, 23:22
6
9
00:00

Difficulty:

65% (hard)

Question Stats:

53% (01:50) correct 47% (02:05) wrong based on 420 sessions

### HideShow timer Statistics

If k is the sum of the digits of integer m, and m=18n, where n is an integer, which of the following must be true?

A. The sum of the digits of m is 9
B. The sum of the digits of k is 9
C. m is a multiple of 2k
D. k is a multiple of 9
E. k is a multiple of 6

Originally posted by superpus07 on 23 Jul 2012, 10:59.
Last edited by Bunuel on 14 Aug 2012, 23:22, edited 1 time in total.
Edited the question.
Intern
Joined: 21 Sep 2010
Posts: 5
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
1
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.
Director
Joined: 27 May 2012
Posts: 901
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 ??
_________________
- Stne
Math Expert
Joined: 02 Sep 2009
Posts: 58429
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

14 Aug 2012, 23:26
1
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.
_________________
Director
Joined: 27 May 2012
Posts: 901
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
_________________
- Stne
Senior Manager
Joined: 23 Oct 2010
Posts: 323
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

15 Aug 2012, 11:12
m=18n means 18*n? when I saw 18n, I thought k=m=1+8+n=9+n and couldnt find any solution
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth
Director
Joined: 22 Mar 2011
Posts: 590
WE: Science (Education)
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

15 Aug 2012, 11:46
1
LalaB wrote:
m=18n means 18*n? when I saw 18n, I thought k=m=1+8+n=9+n and couldnt find any solution

It wasn't stated "the three-digit number 18n".
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
Intern
Joined: 19 Feb 2012
Posts: 7
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

24 Nov 2012, 07:16
I think b is also correct

Posted from GMAT ToolKit
Intern
Joined: 29 Dec 2012
Posts: 3
WE: Information Technology (Computer Software)
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

27 Jan 2013, 11:48
1
B is incorrect. Try m=18 * 11 and you will find that the sum of digits is not 9.

-Abhishek
Manager
Joined: 16 Feb 2012
Posts: 148
Concentration: Finance, Economics
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

12 Feb 2013, 09:28
Could someone explain in which cases k is not a multiple of 6. Thank you!
_________________
Kudos if you like the post!

Failing to plan is planning to fail.
Math Expert
Joined: 02 Sep 2009
Posts: 58429
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

12 Feb 2013, 09:33
Stiv wrote:
If k is the sum of the digits of integer m, and m=18n, where n is an integer, which of the following must be true?

A. The sum of the digits of m is 9
B. The sum of the digits of k is 9
C. m is a multiple of 2k
D. k is a multiple of 9
E. k is a multiple of 6

Could someone explain in which cases k is not a multiple of 6. Thank you!

m=18 --> k=1+8=9 --> 9 is NOT a multiple of 6.
_________________
SVP
Joined: 06 Sep 2013
Posts: 1571
Concentration: Finance
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

12 Nov 2013, 17:23
superpus07 wrote:
If k is the sum of the digits of integer m, and m=18n, where n is an integer, which of the following must be true?

A. The sum of the digits of m is 9
B. The sum of the digits of k is 9
C. m is a multiple of 2k
D. k is a multiple of 9
E. k is a multiple of 6

I could figure out an easy solution for this one. Anyone have any idea how to solve this efficiently?
Will throw some nice Kudos out there!

Cheers
J
Intern
Joined: 24 May 2012
Posts: 5
Concentration: Technology, Entrepreneurship
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

07 Dec 2013, 15:26
Can someone post an example of a case when C is not true?
Retired Moderator
Joined: 16 Jun 2012
Posts: 991
Location: United States
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

08 Dec 2013, 01:57
3
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."

Chris Bangle - Former BMW Chief of Design.
Senior Manager
Joined: 13 Oct 2016
Posts: 359
GPA: 3.98
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

22 Nov 2016, 04:38
1
superpus07 wrote:
If k is the sum of the digits of integer m, and m=18n, where n is an integer, which of the following must be true?

A. The sum of the digits of m is 9
B. The sum of the digits of k is 9
C. m is a multiple of 2k
D. k is a multiple of 9
E. k is a multiple of 6

We can apply the concept of “Digital Root” here.

Digital root is consecutive summation of digits of a number until the sum reaches a one digit value. Although the question is not asking about consecutive summation, all the principles of digital root can still be applied here.

There is ,so called, “Rule of 9”. That is when we multiple any number by 9 its digital root will ALWAYS be 9. Also from the perspective of divisibility rules: when we multiply any number by 9 we’ll make this number a multiple of 9, thus sum of its digit will be divisible by 9.

We have $$m=18n$$ ---> $$m=2n*9$$

Whatever nonzero $$n$$ we plug in the digital root of $$m$$ will ALWAYS be 9.

Now, because we are not asked about digital root directly we need to know another important quality of digital root. Digital root of ANY number has a cycle of 9.

$$9; 18=1+8=9; 27=2+7=9; 36=3+6=9 …. 1152 = 1+1+5+2 = 9 …$$

So we have arithmetic progression of multiples of 9.

Hence if we multiply ANY number by 9, sum of its digit will ALWAYS be a multiple of 9. Even if we plug in n=0, we get m=0 and 0 is multiple of any number.

Director
Joined: 31 Jul 2017
Posts: 512
Location: Malaysia
GMAT 1: 700 Q50 V33
GPA: 3.95
WE: Consulting (Energy and Utilities)
Re: If k is the sum of the digits of integer m, and m=18n, where  [#permalink]

### Show Tags

22 Jan 2018, 09:07
superpus07 wrote:
If k is the sum of the digits of integer m, and m=18n, where n is an integer, which of the following must be true?

A. The sum of the digits of m is 9
B. The sum of the digits of k is 9
C. m is a multiple of 2k
D. k is a multiple of 9
E. k is a multiple of 6

Its an easy problem...lets look at this way.. as per question $$\frac{m}{18} = n$$.. which means $$m = 18p$$ where $$p = 0,1,2,3,4,5,...$$.
Now, K = sum of digits of m, however when m = 0, sum of digits of K will not be equal to 9.
When, m =18.. K will not be multiple of 6...
K will always be multiple of 9.
_________________
If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!
Re: If k is the sum of the digits of integer m, and m=18n, where   [#permalink] 22 Jan 2018, 09:07
Display posts from previous: Sort by