What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is

Question

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

Bunuel · Accepted Answer

30 sec approach:

Given:  9^1+(9^2+9^3+9^4+9^5+9^6+9^7+9^8+9^9) .

Notice that inside the parentheses, we have the sum of 8  odd multiples of 3 . Since the sum of 8 odd numbers is even, the sum inside the parentheses will be an  even multiple of 3 , making it a multiple of 6. Therefore, the entire sum in the parentheses yields a remainder of 0 when divided by 6. We are left with the first term, 9, which gives a remainder of 3 upon division by 6.

Answer: B.

omerrauf · Answer

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 .

Bunuel · Answer

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amitbharadwaj7 · Answer

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.

stonecold · Answer

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

generis · Answer

I used cyclicity, but I'm not sure it's correct, though it gave me the correct answer.

9^1  = 9
 9^2  = 81
 9^3  = ..9
 9^4 = ...1

Odd powers of nine have a units digit of 9, even powers of nine have a units digit of 1.

There are 5 odd powers of nine (1,3,5,7,9). Units digit is 9, 9*5 = 45, so odd powers' sum will have units digit of 5.

There are four even powers of nine(2,4,6,8). Each ends in 1. 4*1 = 4. Even powers' sum will have a units digit of 4.

Add the two groups: 5 + 4 = 9.

9/6 = remainder 3.

Answer B

You could also pair  four  odd powers that end in 9 with four even powers that end in 1: 9,1,9,1,9,1,9,1. Each pair sums to 9 + 1 = 10, with a last digit of 0.

Add one more odd power of 9, which will end in 9 (we have five odd powers, we've only used four). 0 + 9 = 9. 9/6 has R3.

I took 33 seconds to answer. I'm a little nervous.

Bunuel  - are these methods correct?

KrishnakumarKA1 · Answer

Solution:

Given:   9^1+ 9^2+9^3+ …………………………….+9^9

Approach:  Let’s take  N = 9^1+ 9^2+9^3+ …………………………….+9^9

We have to find the remainder when N is divided by 6.

N = 9^1+ (9^2+9^3+ 9^4+ 9^5+ 9^6+ 9^7+ 9^8+9^9) 
If we look at the terms in the brackets we have the sum of 8 odd numbers which are multiples of 3. We have to note here that sum of 8 ODD numbers will result in EVEN.  .
So, the sum in brackets is multiple of 6 and obviously, the remainder will be ZERO. 
So, we are left with the first term 9, the remainder will be “3” when divided by 6.

The correct answer option is “B”.

GMATBusters · Answer

9^1 + 9^2 + 9^3 +...+ 9^9  is odd integer ( as we have 9 odd terms)

(9^1 + 9^2 + 9^3 +...+ 9^9 )/6
= 3k/6, where k is odd

if we divide the numerator and denominator by 3, we will get k/2, where k is odd. 
So, remainder of k/2 = 1

But since,   we have divided the original fraction by 3, the remainder has also been divided by 3  , so actual remainder = 1*3 = 3

Answer B

BrushMyQuant · Answer

We need to find the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

To solve this problem we need to find the cycle of remainder of power of 9 when divided by 6

Remainder of  9^1  (=9) by 6 = 3
Remainder of  9^2  (=81) by 6 = 3
Remainder of  9^3  (=729) by 6 = 3

=> Cycle is 1

=> Remainder of 9^1 + 9^2 + 9^3 +...+ 9^9 by 6 = Remainder of each term by 6
 = 3 + 3 ....+ 3 (9 times) = 3*9 = 27

But remainder cannot be more than 6
=> Remainder = Remainder of 27 by 6 = 3

So,  Answer will be B 
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

https://www.youtube.com/watch?v=h1HTEhmUb1w

bumpbot · Answer

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What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is

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