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 have the formula <math>\frac{a(r^n-1)}{r-1}</math> for the sum of a finite geometric sequence which we want to find the residue modulo 8.
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<cmath>\frac{1 \cdot (3^{2010}-1)}{2}</cmath>
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<cmath>\frac{3^{2010}-1}{2} = \frac{9^{1005}-1}{2}</cmath>
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<cmath>\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8</cmath>
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Therefore, the numerator of the fraction is divisible by <math>8</math>. However, when we divide the numerator by <math>2</math>, we get a remainder of <math>4</math> modulo <math>8</math>, giving us <math>\mathrm{(D)}</math>.
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Note: you need to prove that <cmath>\frac{9^{1005}-1}{2}</cmath> is not congruent to 0 mod 16 because if so, then the whole thing would be congruent to 0 mod 8, even after dividing by 2 ~ ilikepi12
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==Video Solution==
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https://youtu.be/V8VydUpAsS8
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~savannahsolver
  
 
== 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}}
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{{MAA Notice}}

Latest revision as of 19:45, 19 January 2021

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 have the formula $\frac{a(r^n-1)}{r-1}$ for the sum of a finite geometric sequence which we want to find the residue modulo 8. \[\frac{1 \cdot (3^{2010}-1)}{2}\] \[\frac{3^{2010}-1}{2} = \frac{9^{1005}-1}{2}\] \[\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8\] Therefore, the numerator of the fraction is divisible by $8$. However, when we divide the numerator by $2$, we get a remainder of $4$ modulo $8$, giving us $\mathrm{(D)}$.

Note: you need to prove that \[\frac{9^{1005}-1}{2}\] is not congruent to 0 mod 16 because if so, then the whole thing would be congruent to 0 mod 8, even after dividing by 2 ~ ilikepi12

Video Solution

https://youtu.be/V8VydUpAsS8

~savannahsolver

See also

2009 AMC 10B (ProblemsAnswer KeyResources)
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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