2009 AMC 10B Problems/Problem 21
Contents
Problem
What is the remainder when is divided by 8?
Solution
Solution 1
The sum of any four consecutive powers of 3 is divisible by and hence is divisible by 8. Therefore
is divisible by 8. So the required remainder is . The answer is .
Solution 2
We have . Hence for any we have , and then .
Therefore our sum gives the same remainder modulo as . There are terms in the sum, hence there are pairs of , and thus the sum is .
Solution 3
We have the formula for the sum of a finite geometric sequence which we want to find the residue modulo 8. Therefore, the numerator of the fraction is divisible by . However, when we divide the numerator by , we get a remainder of modulo , giving us .
Note: you need to prove that is not congruent to 0 mod 16 because if so, then the whole thing would be congruent to 0 mod 8, even after dividing by 2 ~ ilikepi12
Solution 4
The sum of the sequence is which is equal to . The remainder when 1005 is divided by 8 is 5 If you start dividing the powers of 3 by 8 you will find a pattern rem : 3, rem : 1, rem : 3, rem :1, and so on. All the odd powers of three (positive) have a remainder of when divided by 8, so is going to have remainder of .
Since we know that the remainder of is and that the remainder of is we can substitute it back to our expression
-> , and the remainder when 20 is divided by 8 is 4, . ~LUISFONSECA123
Video Solution
~savannahsolver
See also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AMC 10 Problems and Solutions |
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