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Difference between revisions of "2009 AMC 10B Problems/Problem 21"

(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
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: <math>(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{3009})</math>
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: <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 of <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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== Solution 4 (Patterns) ==
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We can see that <math>\frac{3^0}{8}</math> is has a remainder of 1. <math>\frac{3^1}{8}</math> has a remainder of 4. <math>\frac{3^2}{8}</math> also has a remainder of 4. <math>\frac{3^3}{8}</math> has a remainder of 0.
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 +
If we keep on going, we can see that this pattern repeats every 4 powers of 3 - that is to say, the reminder cycles through <math>1</math>, <math>4</math>, <math>4</math>, and <math>0</math>.
 +
 +
Using simple math, we can see that if assume the pattern repeats every 4 numbers, <math>3^{2009}</math> has a remainder of <math>\boxed {4}</math>. The answer is <math>\mathrm{(D)}</math>.
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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 00:16, 16 October 2024

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 of $1+3$, 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

Solution 4 (Patterns)

We can see that $\frac{3^0}{8}$ is has a remainder of 1. $\frac{3^1}{8}$ has a remainder of 4. $\frac{3^2}{8}$ also has a remainder of 4. $\frac{3^3}{8}$ has a remainder of 0.

If we keep on going, we can see that this pattern repeats every 4 powers of 3 - that is to say, the reminder cycles through $1$, $4$, $4$, and $0$.

Using simple math, we can see that if assume the pattern repeats every 4 numbers, $3^{2009}$ has a remainder of $\boxed {4}$. The answer is $\mathrm{(D)}$.

Video Solution

https://youtu.be/V8VydUpAsS8

~savannahsolver

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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