Difference between revisions of "2009 AMC 10B Problems/Problem 21"
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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>. | 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>. | ||
− | 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 of <math>1+3</math>, and thus the sum is <math>1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8</math>. |
=== Solution 3 === | === Solution 3 === | ||
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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 | 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 | ||
+ | |||
+ | === Solution 4 === | ||
+ | |||
+ | The sum of the sequence is <math>2010\ast</math><math>\frac{3^{2009}+1}{2}</math> which is equal to <math>1005\ast 3^{2009}+1005</math>. | ||
+ | The remainder when 1005 is divided by 8 is 5 | ||
+ | If you start dividing the powers of 3 by 8 you will find a pattern | ||
+ | <math>3^{1}</math> rem : 3, <math>3^{2}</math> rem : 1, <math>3^{3}</math> rem : 3, <math>3^{4}</math> rem :1, and so on. | ||
+ | All the odd powers of three (positive) have a remainder of <math>3</math> when divided by 8, so <math>3^{2009}</math> is going to have remainder of <math>3</math>. | ||
+ | |||
+ | Since we know that the remainder of <math>1005</math> is <math>5</math> and that the remainder of <math>3^{2009}</math> is <math>3</math> we can substitute it back to our expression | ||
+ | |||
+ | <math>1005*3^{2009}+1005</math> -> <math>5*3 + 5 = 20</math>, and the remainder when 20 is divided by 8 is 4, <math>\mathrm{(D)}</math>. ~LUISFONSECA123 | ||
==Video Solution== | ==Video Solution== |
Revision as of 15:17, 8 July 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 of , 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
Solution 4
The sum of the sequence is which is equal to . The remainder when 1005 is divided by 8 is 5 If you start dividing the powers of 3 by 8 you will find a pattern rem : 3, rem : 1, rem : 3, rem :1, and so on. All the odd powers of three (positive) have a remainder of when divided by 8, so is going to have remainder of .
Since we know that the remainder of is and that the remainder of is we can substitute it back to our expression
-> , and the remainder when 20 is divided by 8 is 4, . ~LUISFONSECA123
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.