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 === | ||
+ | |||
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 === | ||
+ | |||
+ | 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>. | ||
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+ | === Solution 3 === | ||
+ | |||
+ | 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. | ||
+ | <cmath>\frac{1 \cdot (3^{2010}-1)}{2}</cmath> | ||
+ | <cmath>\frac{3^{2010}-1}{2} = \frac{9^{1005}-1}{2}</cmath> | ||
+ | <cmath>\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8</cmath> | ||
+ | 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== | ||
+ | https://youtu.be/V8VydUpAsS8 | ||
+ | |||
+ | ~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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:45, 19 January 2021
Contents
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 have the formula for the sum of a finite geometric sequence which we want to find the residue modulo 8. Therefore, the numerator of the fraction is divisible by . However, when we divide the numerator by , we get a remainder of modulo , giving us .
Note: you need to prove that 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
~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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