Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.

DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?

Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.

DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?

Rule for divisibility by 3: The sum of the digits of the number should be multiple of 3

Rule for divisibility by 9: The sum of the nos digits of the number should be divisible by 9 or the number should be divisible by 3 two times.

Rules for divisibility by 27: The sum of the digits should a multiple of 27

Consider no 11111111...27 times = The sum 27*1=27----> divisbible by 3,9 and 27

consider number to be 222....27 times, then sum of the no. 27*2=54 divisibly by 3,9 and 27

So why so because when you sum the numbers either you can add the digits 27 times or multiply the digit *27..

Note that since 27 is divisble by 27,9 and 3 and thus the sum of the nos will be divisible by all the nos.

Re: If n is a 27-digit positive integer, all of whose digits are [#permalink]

Show Tags

09 Aug 2014, 08:36

Wounded Tiger,

Do we have divisibility rule for 27 as "The sum of the digits should a multiple of 27"? Say we have 54 whose sum of digits is 9. Is 9 divisible by 27? Can you explain in detail? Thanks

Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.

DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?

Rule for divisibility by 3: The sum of the digits of the number should be multiple of 3

Rule for divisibility by 9: The sum of the nos digits of the number should be divisible by 9 or the number should be divisible by 3 two times.

Rules for divisibility by 27: The sum of the digits should a multiple of 27

Consider no 11111111...27 times = The sum 27*1=27----> divisbible by 3,9 and 27

consider number to be 222....27 times, then sum of the no. 27*2=54 divisibly by 3,9 and 27

So why so because when you sum the numbers either you can add the digits 27 times or multiply the digit *27..

Note that since 27 is divisble by 27,9 and 3 and thus the sum of the nos will be divisible by all the nos.

Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.

DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?

I think the best solution is provided in spoiler.
_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

Re: If n is a 27-digit positive integer, all of whose digits are [#permalink]

Show Tags

10 Aug 2014, 20:19

sri30kanth wrote:

Wounded Tiger,

Do we have divisibility rule for 27 as "The sum of the digits should a multiple of 27"? Say we have 54 whose sum of digits is 9. Is 9 divisible by 27? Can you explain in detail? Thanks

Hmmm....I agree with smyarga...

Look in general, we know a number is divisible by 3 if the sum of the digits of the nos is divisble by 3

For a number to divisible by 9, either the sum of the nos should be divisble by 9 or the number should be divisble by 3 two times

Similarly, for a no to be divisible by 27, it should be divisble by 3 three times...

There is no other method...

In my approach initially, I just extended the rule for 3 and 9 to 27 and as it can be seen it does not work out...Just went with flow I guess...

Like smyarga said, Best approach is the one given in the spoiler...
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.

DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?

Its not necessary to write the whole number. Though I have taken n=2 in my earlier post, its not necessary as well. We can also solve it by variable

nnn...........n ......... 27 times

Addition of digits = n+n+n+ ........... 27 times = 27n

Re: If n is a 27-digit positive integer, all of whose digits are [#permalink]

Show Tags

31 Oct 2016, 19:59

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Suppose n = 111,111,111,111,111,111,111,111,111. Then the digits of n sum up to 27 * 1, which is divisible by 9, so n is divisible by both 3 and 9 by the basic divisibility rules given in our Arithmetic book. (Any other 27-digit integer with all identical digits is just a multiple of this first one, so we know that all the other values of n will divide by 3 and 9.) We might conclude that the divisibility rule "generalizes", and that any number whose digits sum to a multiple of 27 divides by 27. Be careful, though! This is exact the sort of natural, intuitive thinking that the testwriters like to exploit, so let's see if we can test a number. Suppose n = 111,111,111,111,111,111,111,111,111. Noticing that 111/3 = 37, we can divide n by 3 and obtain n/3 = 37,037,037,037,037,037,037,037,037. The sum of the digits of this number is 90, so n/3 divides by 9. (n/3)/9 = n/27, so n divides by 27. Since any 27-digit number with one repeating digit is just a multiple of 111,111,111,111,111,111,111,111,111 -- i.e., a multiple of a multiple of 27 -- we conclude that n is divisible by 27.

DO WE HAVE TO WRITE OUT THE WHOLE NUMBER? IS THERE ANOTHER WAY TO SOLVE? ALSO WHY DO WE DIVIDE 111 by 3? AND THEN HOW DO WE GET TO n/3 = 37,037,037,037,037,037,037,037,037?

\(N=aaaaaaaaaaaaa…\). (27 times)

\(a+a+a+a+a+ … (27 times) = 27*a\)

Number is definitely divisible by 3 and 9. I and II are correct.

With 27 things are a bit trickier.

\(27*37=999\) which is \(10^3 – 1\)

When we need to find remainder of a number in a form \(10^x – 1\) we need to take clusters of digits taken by \(x\) starting from the right and sum them up.

In our case \(aaaaaa… 27\) times we split the number into the groups of \(aaa\) starting from the right. We’ll get:

\(\frac{27}{3} *(aaa) = 9*aaa\)

Because we have particular case where all digits are the same, our group of aaa is divisible by 3. We can put

\(9*3x = 27x\) and our number is divisible by 27. Option III is also correct.

Answer E

In GENERAL divisibility by 3 and 9 does not automatically mean that number is divisible by 27. As example: 13779 (is divisible by 9 but not by 27). And we can't apply divisibility by 9 rule here. Although the sum of the diggits is 27 the number is not divisible by 27!

gmatclubot

If n is a 27-digit positive integer, all of whose digits are
[#permalink]
12 Dec 2016, 03:16

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...