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

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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."

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]

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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]

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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 ??
_________________

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

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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]

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19 Jun 2015, 08:42

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