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

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

Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink]
14 Aug 2012, 22:26

1

This post received KUDOS

Expert's post

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).

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

Re: If k is the sum of the digits of integer m, and m=18n, where [#permalink]
08 Dec 2013, 00:57

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.

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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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