BTGmoderatorDC wrote: ↑Tue Apr 21, 2020 5:35 pm
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
OA
E
Source: Veritas Prep
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
