Difference between revisions of "2009 AMC 10B Problems/Problem 21"

Revision as of 15:16, 8 July 2021

Problem

What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?

$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$

Solution

Solution 1

The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore

$(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{2009})$

is divisible by 8. So the required remainder is $3^0 + 3^1 = \boxed {4}$ . The answer is $\mathrm{(D)}$ .

Solution 2

We have $3^2 = 9 \equiv 1 \pmod 8$ . Hence for any $k$ we have $3^{2k}\equiv 1^k = 1 \pmod 8$ , and then $3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3 \pmod 8$ .

Therefore our sum gives the same remainder modulo $8$ as $1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3$ . There are $2010$ terms in the sum, hence there are $2010/2 = 1005$ pairs of $1+3$ , and thus the sum is $1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8$ .

Solution 3

We have the formula $\frac{a(r^n-1)}{r-1}$ for the sum of a finite geometric sequence which we want to find the residue modulo 8. $\frac{1 \cdot (3^{2010}-1)}{2}$ $\frac{3^{2010}-1}{2} = \frac{9^{1005}-1}{2}$ $\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8$ Therefore, the numerator of the fraction is divisible by $8$ . However, when we divide the numerator by $2$ , we get a remainder of $4$ modulo $8$ , giving us $\mathrm{(D)}$ .

Note: you need to prove that $\frac{9^{1005}-1}{2}$ 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 $2010\ast$ $\frac{3^{2009}+1}{2}$ which is equal to $1005\ast 3^{2009}+1005$ . 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 $3^{1}$ rem : 3, $3^{2}$ rem : 1, $3^{3}$ rem : 3, $3^{4}$ rem :1, and so on. All the odd powers of three (positive) have a remainder of $3$ when divided by 8, so $3^{2009}$ is going to have remainder of $3$ .

Since we know that the remainder of $1005$ is $5$ and that the remainder of $3^{2009}$ is $3$ we can substitute it back to our expression

$1005*3^{2009}+1005$ -> 5*3 + 5 = 20, and the remainder when 20 is divided by 8 is 4, $\mathrm{(D)}$ .

Video Solution

https://youtu.be/V8VydUpAsS8

~savannahsolver

@@ 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
+=== 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> -> 5*3 + 5 = 20, and the remainder when 20 is divided by 8 is 4, <math>\mathrm{(D)}</math>.
 ==Video Solution==

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
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