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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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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 ??
_________________

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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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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08 Dec 2013, 01:57

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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.
_________________

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]

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19 Jun 2015, 08:42

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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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22 Nov 2016, 04:38

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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.

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.

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