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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>.
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===Solution 3===
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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>
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Then 1-1=0mod 8$
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-Williamgolly
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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 11:24, 9 December 2019

Problem

What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?

$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$

Solution

Solution 1

The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore

$(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{2009})$

is divisible by 8. So the required remainder is $3^0 + 3^1 = \boxed {4}$. The answer is $\mathrm{(D)}$.

Solution 2

We have $3^2 = 9 \equiv 1 \pmod 8$. Hence for any $k$ we have $3^{2k}\equiv 1^k = 1 \pmod 8$, and then $3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3 \pmod 8$.

Therefore our sum gives the same remainder modulo $8$ as $1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3$. There are $2010$ terms in the sum, hence there are $2010/2 = 1005$ pairs $1+3$, and thus the sum is $1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8$.


Solution 3

We see the sum is $3^2010-1$ from base number arithmetic. $3^2010=(3^2)^1005=1^1005=1 mod 8$ 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

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