2009 AMC 10B Problems/Problem 21
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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 
, and thus the sum is 
.
See also
| 2009 AMC 10B (Problems • Resources) | ||
| Preceded by Problem 20 | Followed by Problem 22 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
