M7MBA wrote: ↑Sat Mar 10, 2018 3:00 am
If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?
I. n is divisible by 3
II. n is divisible by 9
III. n is divisible by 27
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III
The OA is the option
E.
Experts, may you give me some help here? I'd be thankful.
We can let n = 111,111,111,111,111,111,111,111,111. If n is divisible by 3, 9, and 27, then kn (where k = 2, 3, …, 9) will be also divisible by 3, 9, and 27, respectively (notice that kn is any one of the other 27-digit numbers whose digits are the same).
We see that the sum of the digits of n is 27. Since 27 is a multiple of both 3 and 9, we see that n is divisible by both 3 and 9 (recall that the rules for divisibility of 3 and 9 is that the sum of the digits of the number has to be a multiple of 3 and 9, respectively).
However, in order for n to be divisible by 27, the quotient of n/3 has to be divisible by 9, or the quotient of n/9 has to be divisible by 3. Let’s verify the former.
Since 111/3 = 37, so n/3 = 37,037,037,037,037,037,037,037,037. We see that there are 9 groups of 37 (or 037), so the sum of the digits of n/3 is (3 + 7) x 9 = 90. Since 90 is a multiple of 9, n/3 is divisible by 27. In other words, n is divisible by 27.
Answer: E
