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
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?
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