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

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If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 23 Jul 2012, 09:59
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Aug 2012, 22:22, edited 1 time in total.
Edited the question.
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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 23 Jul 2012, 11:48
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.
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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 14 Aug 2012, 22:10
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 ??
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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 14 Aug 2012, 22:26
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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.
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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 14 Aug 2012, 22:34
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
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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 15 Aug 2012, 10:12
m=18n means 18*n? when I saw 18n, I thought k=m=1+8+n=9+n and couldnt find any solution
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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 15 Aug 2012, 10:46
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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".
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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 24 Nov 2012, 06:16
I think b is also correct

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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 27 Jan 2013, 10:48
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B is incorrect. Try m=18 * 11 and you will find that the sum of digits is not 9.

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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 12 Feb 2013, 08:28
Could someone explain in which cases k is not a multiple of 6. Thank you!
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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 12 Feb 2013, 08:33
Expert's post
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.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 12 Nov 2013, 16:23
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!

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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 07 Dec 2013, 14:26
Can someone post an example of a case when C is not true?
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Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink] New post 08 Dec 2013, 00:57
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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.
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Re: If k is the sum of the digits of integer m, and m=18n, where   [#permalink] 08 Dec 2013, 00:57
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