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

Question

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

IanStewart · Answer

We have a 27 digit number with 27 identical digits. Let's say 'A' is that digit. Look at the number formed by just the last nine digits first, so let k = AAAAAAAAA (nine digits). The sum of the digits of k is 9A, so the sum of k's digits is divisible by 9, and using the familiar divisibility test, k must be divisible by 9. Now our 27 digit number is just equal to 10^18 * k + 10^9 * k + k. Factoring, our number equals

k (10^18 + 10^9 + 1)

and the number in brackets is just going to look something like this (I'm not necessarily counting the zeros correctly) : 1000000001000000001, so the sum of its digits is 3, and it will be divisible by 3. So k is divisible by 9, and the number in brackets is divisible by 3, so their product is divisible by 27. But their product is the number we're asked about, so our number is divisible by 27, and thus also by 9 and by 3, and the answer is I, II and III.

sujoykrdatta · Answer

Our number is, for example: 
555555... (27 times)
= 5*(11111... 27 times)

Let us focus on the number 
11111... 27 times

Let us check some similar numbers:

111111 (6 times i.e. 2 groups of 111) = 111*(1000+1)
= 111*1001
--- note there are 2 1s in the number inside brackets

111111111 (9times i.e. 3 groups of 111) = 111*1000²+111*1000+111*1
= 111*(1000²+1000+1) = 111*(1001001)
--- note there are 3 1s in the number inside brackets

Thus: 11111... 27 times (i.e. 9 groups of 111)
= 111*(1001001....001) 
--- there would be 9 1s in the number inside brackets

Thus: (1001001....001) is divisible by 9 since sum of digits is 9
Also, 111 is divisible by 3 since sum of digits is divisible by 3

Thus, the given number is divisible by 9*3 = 27

Thus, all three statements are correct

Answer E

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cathygeorge123 · Answer

Guys, please learn up your divisibility rules from the GMATCLUB Math Book.

For 3 & 9 : If the sum of the digits is divisible by 3 or 9, so is the number divisible by 3 or 9

Let's have the 27 digit number = x x x x x......i.e 27 x's 
Hence Sum = 27 x

So 27x divisible by 3 and 9. So (I) and (II) are OK
If 27x divisible by 3 and 9, it surely is divisible by 27 ( infact it it enough for us to know that 27x is divisible by 3)
So (III) is also OK

(I), (II), (III) are OK

Ans: (E)

IanStewart · Answer

It is true that the sum of the digits, 27x, is divisible by 27. But there is no divisibility rule that says "if the sum of a number's digits is divisible by 27, the number itself is divisible by 27". That rule would not be true -- for example, 1989 is not divisible by 27, even though the digits of 1989 sum to 27. That's why I and sujoykrdatta above needed to do a bit of work to prove that III is true in this question.

ScottTargetTestPrep · Answer

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

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