Difference between revisions of "2009 AMC 10B Problems/Problem 21"
VelaDabant (talk | contribs) (New page: == Problem == What is the remainder when <math>3^0 + 3^1 + 3^2 + \cdots + 3^{2009}</math> is divided by 8? <math>\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{...) |
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== Solution == | == Solution == | ||
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+ | === Solution 1 === | ||
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The sum of any four consecutive powers of 3 is divisible by <math>3^0 + 3^1 + 3^2 +3^3 = 40</math> and hence is divisible by 8. Therefore | The sum of any four consecutive powers of 3 is divisible by <math>3^0 + 3^1 + 3^2 +3^3 = 40</math> and hence is divisible by 8. Therefore | ||
− | : <math>(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{ | + | : <math>(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{2009})</math> |
is divisible by 8. So the required remainder is <math>3^0 + 3^1 = \boxed {4}</math>. The answer is <math>\mathrm{(D)}</math>. | is divisible by 8. So the required remainder is <math>3^0 + 3^1 = \boxed {4}</math>. The answer is <math>\mathrm{(D)}</math>. | ||
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+ | === Solution 2 === | ||
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+ | We have <math>3^2 = 9 \equiv 1 \pmod 8</math>. Hence for any <math>k</math> we have <math>3^{2k}\equiv 1^k = 1 \pmod 8</math>, and then <math>3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3 \pmod 8</math>. | ||
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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>. | ||
== 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}} |
Revision as of 13:41, 8 March 2009
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 • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AMC 10 Problems and Solutions |