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

Question

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

Jitu20 · Answer

Take numbers for m, n. Let n=1,2,3... for any value of n, the sum of the digits of M is a multiple of 9. 
 
Therefore k is a multiple of 9

ebear444 · Answer

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.

BlackHawk2478 · Answer

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

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Chitroll · Answer

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.

winterschool · Answer

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

Answer D

Kinshook · Answer

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

sujoykrdatta · Answer

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

Answer D

Mapat · Answer

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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Bunuel · Answer

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.

pappal · Answer

if we take n=0,1,2,3,4,5
m=0,18,36,54,72,90
k=0,9,9,9,9,9
2k=0,18,18,18,18,18
so isn't m the multiple of 2k?? not 9 as 0 is not the multiple of 9.
even if we take n=11,111,111,111
m=199,999,999,998
k=99 & 2k=198=2*9*11
m here is also a multiple of 2k.
advance thanks for the correction.

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

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My Rewards