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If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?

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If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?

by BTGmoderatorDC » Tue Apr 21, 2020 5:35 pm

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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

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Re: If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?

by Jay@ManhattanReview » Wed Apr 22, 2020 3:42 am
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
Say the only and the common digit is x; thus, n = xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Thus, n = x(1111111111111111111111111111)

n = x*sum of digits = 27x

We see that 27x is divisible by 3, 9 and 27; thus, the correct Option is E.

The correct answer: E

Hope this helps!

-Jay
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Re: If n is a 27-digit positive integer, all of whose digits are the same, which of the following must be true?

by Scott@TargetTestPrep » Mon Apr 27, 2020 3:19 am
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

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