Difference between revisions of "2009 AMC 10B Problems/Problem 21"
(→Solution 2) |
(→Solution 4) |
(3 intermediate revisions by the same user not shown) | |
(No difference)
|
Latest revision as of 01:22, 6 December 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
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.