Mock AIME 4 2006-2007 Problems/Problem 10

Problem

Compute the remainder when

${2007 \choose 0} + {2007 \choose 3} + \cdots + {2007 \choose 2007}$

is divided by 1000.

Solution

Let $\omega$ and $\zeta$ be the two complex third-roots of 1. Then let

$S = (1 + \omega)^{2007} + (1 + \zeta)^{2007} + (1 + 1)^{2007} = \sum_{i = 0}^{2007} {2007 \choose i}(\omega^i + \zeta^i + 1)$ .

Now, if $i$ is a multiple of 3, $\omega^i + \zeta^i + 1 = 1 + 1 + 1 = 3$ . If $i$ is one more than a multiple of 3, $\omega^i + \zeta^i + 1 = \omega + \zeta + 1 = 0$ . If $i$ is two more than a multiple of 3, $\omega^i + \zeta^i + 1 = \omega^2 + \zeta^2 + 1= \zeta + \omega + 1 = 0$ . Thus

$S = \sum_{i = 0}^{669} 3 {2007 \choose 3i}$ , which is exactly three times our desired expression.

We also have an alternative method for calculating $S$ : we know that $\{\omega, \zeta\} = \{-\frac{1}{2} + \frac{\sqrt 3}{2}i, -\frac{1}{2} - \frac{\sqrt 3}{2}i\}$ , so $\{1 + \omega, 1 + \zeta\} = \{\frac{1}{2} + \frac{\sqrt 3}{2}i, \frac{1}{2} - \frac{\sqrt 3}{2}i\}$ . Note that these two numbers are both cube roots of -1, so $S = (1 + \omega)^{2007} + (1 + \zeta)^{2007} + (1 + 1)^{2007} = (-1)^{669} + (-1)^{669} + 2^{2007} = 2^{2007} - 2$ .

Thus, the problem is reduced to calculating $2^{2007} - 2 \pmod{1000}$ . $2^{2007} \equiv 0 \pmod{8}$ , so we need to find $2^{2007} \pmod{125}$ and then use the Chinese Remainder Theorem. Since $\phi (125) = 100$ , by Euler's Totient Theorem $2^{20 \cdot 100 + 7} \equiv 2^7 \equiv 3 \pmod{125}$ . Combining, we have $2^{2007} \equiv 128 \pmod{1000}$ , and so $3S \equiv 128-2 \pmod{1000} \Rightarrow S\equiv \boxed{042}\pmod{1000}$ .

Solution 2

$\sum_{k=0}^{n} {n \choose k} =2^n$ , and $\sum_{k=0}^{3n} {3n \choose k} =2^{3n}$

$\sum_{k=0}^{3n} {3n \choose 3k}=\frac{\sum_{k=0}^{3n} {3n \choose k}+q(n)}{3}$ where $q(n)$ is an integer $-2 \le q(n) \le 2$ that depends on the value of $n$ and will make the sum an integer. The division by 3 comes from the fact that we're skipping 2 out of every 3 terms in the binomial. So we divide the whole sum by 3 and we add or subtract $q(n)$ to correct for the integer based on the modularity of the sum with 3

$\sum_{k=0}^{3n} {3n \choose 3k}=\frac{2^{3n}+q(n)}{3}$

When $n$ is odd, $3n$ is odd, $2^{odd} \equiv (-1)^{odd}\;(mod\;3)\equiv -1\;(mod\;3)\equiv 2\;(mod\;3)$ , Therefore $q(n)=-2$ because we need to subract 2.

When $n$ is even, $3n$ is even, $2^{even} \equiv (1)^{even}\;(mod\;3)\equiv 1\;(mod\;3)\equiv -2\;(mod\;3)$ , Therefore $q(n)=2$ because we need to add 2.

So, the equation becomes:

$\sum_{k=0}^{3n} {3n \choose 3k}=\frac{2^{3n}+(-1)^n2}{3}$

${2007 \choose 0} + {2007 \choose 3} + \cdots + {2007 \choose 2007}=\sum_{k=0}^{3\times 669} { 3\times 669 \choose 3k }=\frac{2^{2007}+(-1)^{669}2}{3}=\frac{2^{2007}-2}{3}$

$2^{2007} \equiv 2^7\;(mod\;1000)\equiv 128\;(mod\;1000)$

$2^{2007}-2 \equiv 128-2\;(mod\;1000)\equiv 126\;(mod\;1000)$

$\frac{2^{2007}-2}{3} \equiv \frac{126}{3}\;(mod\;1000)\equiv 42\;(mod\;1000)$

Therefore, the remainder when ${2007 \choose 0} + {2007 \choose 3} + \cdots + {2007 \choose 2007}$ is divided by 1000 is $\boxed{042}$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Mock AIME 4 2006-2007 (Problems, Source)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Page

Toolbox

Search

Mock AIME 4 2006-2007 Problems/Problem 10

Contents

Problem

Solution

Solution 2

See also