2004 AIME II Problems/Problem 10

Problem

Let $S$ be the set of integers between 1 and $2^{40}$ whose binary expansions have exactly two 1's. If a number is chosen at random from $S,$ the probability that it is divisible by 9 is $p/q,$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution

A positive integer $n$ has exactly two 1s in its binary representation exactly when $n = 2^j + 2^k$ for $j \neq k$ nonnegative integers. Thus, the set $S$ is equal to the set $\{n \in \mathbb{Z} \mid n = 2^j + 2^k \,\mathrm{ and }\, 0 \leq j < k \leq 39\}$ . (The second condition ensures simultaneously that $j \neq k$ and that each such number less than $2^{40}$ is counted exactly once.) This means there are ${40 \choose 2} = 780$ total such numbers.

Now, consider the powers of 2 mod 9: $2^{6n} \equiv 1, 2^{6n + 1} \equiv 2, 2^{6n + 2} \equiv 4, 2^{6n + 3} \equiv 8 \equiv -1, 2^{6n + 4} \equiv 7 \equiv -2, 2^{6n + 5} \equiv 5 \equiv -4 \pmod 9$ .

It's clear what the pairs $j, k$ can look like. If one is of the form $6n$ (7 choices), the other must be of the form $6n + 3$ (7 choices). If one is of the form $6n + 1$ (7 choices) the other must be of the form $6n + 4$ (6 choices). And if one is of the form $6n + 2$ (7 choices), the other must be of the form $6n + 5$ (6 choices). This means that there are $7\cdot 7 + 7\cdot 6 + 7\cdot 6 = 49 + 42 +42 = 133$ total "good" numbers.

So our probability is $\frac{133}{780}$ and our answer is $133 + 780 = 913$ .

Page

Toolbox

Search

2004 AIME II Problems/Problem 10

Problem

Solution

See also