If n is a 27-digit positive integer, all of whose digits

[M7MBA]

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.

[GMATGuruNY]

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

An integer is a multiple of 9 if its digit sum is a multiple of 9.

Test the smallest possible case:
Let n = 111, 111,111...111,111,111, where all 27 digits are 1.

Since there are 27 digits of 1, the sum of the digits = 27*1 = 27.
Since the digit sum is a multiple of 9, n is divisible by 9.
Since n is divisible by 9, it must also be divisible by 3.

Since 111/3 = 037, dividing n by 3 will yield the following:
37, 037, 037...037, 037, 037.
Here, there will be 9 sets of 037, with the result that the integer in blue will have the following digit sum:
9*37.
Since the digit sum is a multiple of 9, the integer in blue is divisible by 9.
Thus:
n can be divided by 3 and then by 9, implying that n is divisible by 27.

Since the smallest option for n is divisible by 3, 9 and 27, any larger option for n will also be divisible by 3, 9 and 27.

The correct answer is E.

[Scott@TargetTestPrep]

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