Number theory -4

guerrero 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

OA[spoiler]D

Having trouble understanding the problem .. [/spoiler]

If m = 18n, then m is a multiple of 18.
In other words, m could equal ...-36, -18, 0, 18, 36, 54, 72, . . . etc.

Aside: An official GMAT would likely restrict the values of m such that m > 0.

k equals the sum of the digits of integer m

Let's take the possible values of m (...-36, -18, 0, 18, etc.) and find the sums of these integers. We get: 9, 9, 0, 9, 9, etc.
These sums are the possible values of k.
Notice that all possible values of k are divisible by 9.
This should come as no surprise, because if we take all multiples of 9, the sum of their digits is always divisible by 9. (this is a rule)

So, as we can see, the answer is D

Cheers,
Brent

srcc25anu wrote:As Brent pointed out, if m was restricted to be greater than 0, then range of values for m would have been 18, 36, 54, 72.... The sum of digits in each case = 9.
Then even condition B would have been true that sum of digits of K = 9.

+ condition D would stay correct that K is a multiple of 9.

Am I correct?

Yes, you're correct. Although I should point out that, even without the restriction, k is still a multiple of 9. That is, 0 is a multiple of 9.

Cheers,
Brent

guerrero 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?

It is GMAT and only one option among the given will be correct.
So, for this problem, if we somehow identify one choice that is always true, we don't need to check whether others are true or not.

Now, m = 18n = multiple of 9
So, sum digits of m is multiple of 9
--> k is multiple of 9.
--> This must be true always as we haven't assumed anything.

The correct answer is D.