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

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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'.

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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

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?

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. [ODD + ODD = 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”.
 

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

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