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.

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:

Re: What is the remainder when 9^1 + 9^2 + 9^3 +....+ 9^9 is [#permalink]
20 Feb 2012, 22:40

4

This post received KUDOS

Expert's post

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.

Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]
21 Feb 2012, 12:50

4

This post received KUDOS

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]
12 Jul 2013, 02:45

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]
06 Aug 2013, 09:10

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]
06 Aug 2013, 10:10

1

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]
14 Aug 2014, 00:49

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. _________________