Difference between revisions of "2009 AMC 10B Problems/Problem 21"
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===Solution 3=== | ===Solution 3=== | ||
| − | We see the sum is <math>3^2010-1</math> from base number arithmetic. <math>3^2010=(3^2)^1005=1^1005=1 mod 8</math> | + | We see the sum is <math>3^{2010}-1</math> from base number arithmetic. <math>3^{2010}=(3^2)^{1005}=1^{1005}=1 mod 8</math> |
Then 1-1=0mod 8$ | Then 1-1=0mod 8$ | ||
-Williamgolly | -Williamgolly | ||
Revision as of 11:24, 9 December 2019
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 
.
Solution 3
We see the sum is 
from base number arithmetic. 
Then 1-1=0mod 8$
-Williamgolly
I sillied pls correct me
See also
| 2009 AMC 10B (Problems • Answer Key • 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
