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

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Re: If k is the sum of the digits of integer m, and m=18n, where n is an [#permalink]
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ebear444 wrote:
I understand that K is a multiple of 9, but how is A wrong? For every value of n multiplied by 18 is also a multiple of 18 and therefore a multiple of 9.

Take e.g. n=11
this is why A is wrong

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Re: If k is the sum of the digits of integer m, and m=18n, where n is an [#permalink]
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ebear444 wrote:
I understand that K is a multiple of 9, but how is A wrong? For every value of n multiplied by 18 is also a multiple of 18 and therefore a multiple of 9.

A cannot be true because the question says that the value of "n" can be an integer. We all know that 0 is an integer. Hence in that specific case, the value of k is also going to be 0. Hence we can eliminate A.
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Re: If k is the sum of the digits of integer m, and m=18n, where n is an [#permalink]
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

m=18n, n = 1,2,3,...........................
n= 1, m = 18, k= 1+8=9
n=2, m= 36, k=6+3=9
.......
n=55, m= 990, k= 9+9=9*2

so , k is a multiple of 9

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

m = 18; k = 9
m = 36; k = 9
m = 54; k = 9
m = 72; k = 9
m = 90; k = 9
m = 108; k = 9
m = 126; k = 9
m = 144; k = 9
m = 162; k = 9
m = 180; k = 9
m = 198; k = 18

k is a multiple of 9

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

Since m is a multiple of 18, it must be a multiple of 9.
Since m is divisible by 9, the sum of digits of m, i.e., k, must be divisible by 9 (divisibility rule of 9)

The reason for this rule is:
Any number, say, of 3 digits a, b and c can be written as 100a + 10b + c = 99a + 9b + (a + b + c)
Since 99a + 9b is surely divisible by 9, we just need to check the divisibility of (a + b + c), i.e., the sum of digits

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Re: If k is the sum of the digits of integer m, and m=18n, where n is an [#permalink]
Can anybody explain why B is wrong?
18*2=36
Then K=9
Sum of digits of K is 9
18*102=1836
Then K=18
Sum of digits of K is again 9

Then how come B true. I understand the logic behind D being the answer.

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Re: If k is the sum of the digits of integer m, and m=18n, where n is an [#permalink]
Mapat wrote:
Can anybody explain why B is wrong?
18*2=36
Then K=9
Sum of digits of K is 9
18*102=1836
Then K=18
Sum of digits of K is again 9

Then how come B true. I understand the logic behind D being the answer.

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The sum of the digits of k is not necessarily 9. For example, if m = 18*11,111,111,111 = 199,999,999,998, then k = 99 and the sum of the digits of k is 18, not 9.
Re: If k is the sum of the digits of integer m, and m=18n, where n is an [#permalink]
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