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:

What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

A. 0 B. 3 C. 2 D. 5 E. None of the above

30 sec approach: Given: \(9^1+(9^2+9^3+9^4+9^5+9^6+9^7+9^8+9^9)\). Notice that in the brackets we have the sum of 8 odd multiples of 3, hence the sum in the brackets will be even multiple of 3 (the sum of 8 odd numbers is even). So, the sum in the brackets is multiple of 6 (remainder is zero). So we are just left with the first term 9, which yields remainder of 3 upon division by 6.

Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

Show Tags

21 Feb 2012, 13:50

10

This post received KUDOS

5

This post was BOOKMARKED

Don't really know if my approach is correct but this is how I approached it.

When divided by \(6\), \(9^1\) leaves a remainder of \(3\) When divided by \(6\), \(9^2\) leaves a remainder of \(3\) When divided by \(6\), \(9^3\) leaves a remainder of \(3\)

You can check further if you want to, but at this point I had decided that all the terms individually leave a remainder of \(3\), so all the remainder added up would be \(9*3=27\) , and \(27\) divided by \(6\) leaves a remainder of \(3\) . Hence the answer should be B.

If I am correct, remainders can be added and then divided by the original number to come up with the remainder. For example, lets take two numbers, \(11\) and \(13\) and divide them by \(4\). \(11\) and \(13\) add up to \(24\) and \(24\) divided by \(4\) leaves a remainder of \(0\). \(11\) divided by \(4\) leaves a remainder of \(3\), \(13\) divided by \(4\) leaves a remainder of \(1\). Now when you add the remainders, \(3+1=4\), which leaves a remainder of 0 when divided by \(4\) or is divisible by \(4\).
_________________

"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde

Re: What is the remainder when 9^1 + 9^2 + 9^3 +....+ 9^9 is [#permalink]

Show Tags

12 Jul 2013, 03:45

Bunuel wrote:

sunniboy007 wrote:

What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

A. 0 B. 3 C. 2 D. 5 E. None of the above

30 sec approach: Given: \(9^1+(9^2+9^3+9^4+9^5+9^6+9^7+9^8+9^9)\). Notice that in the brackets we have the sum of 8 odd multiples of 3, hence the sum in the brackets will be even multiple of 3 (the sum of 8 odd numbers is even). So, the sum in the brackets is multiple of 6 (remainder is zero). So we are just left with the first term 9, which yields remainder of 3 upon division by 6.

Answer: B.

Hi Bunnel,

I did it as below: Sum = 9/8*(9^9-1) Rem (s/6) = ?

Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

Show Tags

06 Aug 2013, 10:10

1

This post received KUDOS

I searched for the patrons in the digit of nine, which resulted in 1,9,1,9,1,9..... after that I summed them up which was 49. 49 divided by 6 left a remainder of 3.

Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

Show Tags

06 Aug 2013, 11:10

5

This post received KUDOS

6 is an even multiple of 3. When any even multiple of 3 is divided by 6, it will leave a remainder of 0. Or in other words it is perfectly divisible by 6.

On the contrary, when any odd multiple of 3 is divided by 6, it will leave a remainder of 3. For e.g when 9 an odd multiple of 3 is divided by 6, you will get a remainder of 3.

9 is an odd multiple of 3. And all powers of 9 are odd multiples of 3. Therefore, when each of the 9 powers of 9 listed above are divided by 6, each of them will leave a remainder of 3.

The total value of the remainder = 3 + 3 + .... + 3 (9 remainders) = 27. 27 is divisible by 6. Hence, it will leave remainder as 3.

Hence, the final remainder when the expression 9^1 + 9^2 + 9^3 + .... + 9^9 is divided by 6 will be equal to '3'. and one more point to add if the expression is 9^1+9^2+...........+9^10 is divided by 6 then the remainter will be '0'

We can generalize it further:- if (9^1+9^2+.......9^n) if n is odd then the remainder will always be 3 and if n is even then the remainder will always be '0'.

I hope people will like this explaination and if it helps you further please give Kudos to me.

Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

Show Tags

14 Aug 2014, 01:49

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.
_________________

Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

Show Tags

14 Mar 2016, 00:54

Here The key o such kind of questions is to find any pattern here sum of odd terms yields remainder of 3 and sum of even => remainder =0 since there are 9 terms involved => remainder = 3 Hence B
_________________

Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

Show Tags

17 Mar 2016, 05:42

1

This post received KUDOS

The easiest way here is to find he pattern here 9^1/6=> remainder =3 9^1+9^2/6=> reminder = 0 9^1+9^2+9^3/6=> remainder =3 hence the cyclicity is 2 so the number of terms are odd => remainder =3 hence B
_________________

Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

Show Tags

31 Mar 2016, 10:20

An arithmetics question here, isn't 9^1 + 9^2 + 9^3 +...+ 9^9 the same as 9^11? like factor all the common nines, it will give you 9^2 (nine nines) then add them to the given 9^9 and get 9^11. Does this make sense?... We are given a sum, so Im not sure this logic works..

An arithmetics question here, isn't 9^1 + 9^2 + 9^3 +...+ 9^9 the same as 9^11? like factor all the common nines, it will give you 9^2 (nine nines) then add them to the given 9^9 and get 9^11. Does this make sense?... We are given a sum, so Im not sure this logic works..

Thank you!

No, this does not make sense. Not sure how you are getting this... You CANNOT factor out 9^2 out of 9^1 + 9^2 + 9^3 +...+ 9^9, you can only factor out 9.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...