Difference between revisions of "2009 AMC 10B Problems/Problem 21"
(→Solution 4) |
|||
| (4 intermediate revisions by the same user not shown) | |||
| Line 31: | Line 31: | ||
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 (Patterns) == | ||
| + | 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. | ||
| + | |||
| + | 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>. | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 00:16, 16 October 2024
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.
![\[\frac{1 \cdot (3^{2010}-1)}{2}\]](http://latex.artofproblemsolving.com/6/d/b/6db10a23c131768960b3f0f57e141d4e9d2edcec.png)
![\[\frac{3^{2010}-1}{2} = \frac{9^{1005}-1}{2}\]](http://latex.artofproblemsolving.com/d/0/5/d05bd1b955c4f2eb083e52b012d3d33c86a94702.png)
![\[\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8\]](http://latex.artofproblemsolving.com/6/8/a/68a6dafa6a361a53ef0f2421e265eed91901085a.png)
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 ![\[\frac{9^{1005}-1}{2}\]](http://latex.artofproblemsolving.com/b/9/e/b9e6e42c3fae1e6d2f6a9b861c9e974a79a5cbd4.png)
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 
is has a remainder of 1. 
has a remainder of 4. 
also has a remainder of 4. 
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 
, , , and 
.
Using simple math, we can see that if assume the pattern repeats every 4 numbers, 
has a remainder of 
. The answer is .
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. 
