Difference between revisions of "2009 AMC 10B Problems/Problem 21"
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Therefore our sum gives the same remainder modulo <math>8</math> as <math>1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3</math>. There are <math>2010</math> terms in the sum, hence there are <math>2010/2 = 1005</math> pairs <math>1+3</math>, and thus the sum is <math>1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8</math>. | Therefore our sum gives the same remainder modulo <math>8</math> as <math>1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3</math>. There are <math>2010</math> terms in the sum, hence there are <math>2010/2 = 1005</math> pairs <math>1+3</math>, and thus the sum is <math>1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8</math>. | ||
+ | |||
+ | |||
+ | ===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> | ||
+ | Then 1-1=0mod 8$ | ||
+ | -Williamgolly | ||
+ | |||
+ | I sillied pls correct me | ||
== See also == | == See also == | ||
{{AMC10 box|year=2009|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2009|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:24, 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.