Difference between revisions of "2009 AMC 10B Problems/Problem 21"
(→Solution) |
(→Solution) |
||
Line 42: | Line 42: | ||
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 | 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> -> 5*3 + 5 = 20, and the remainder when 20 is divided by 8 is 4, <math>\mathrm{(D)}</math>. ~LUISFONSECA123 | + | <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 16:17, 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.