# Difference between revisions of "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 1

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.

The probability is $\frac{133}{780}$, and the answer is $133 + 780 = \boxed{913}$.

## Solution 2

Note that $2^3 \equiv -1\text{ (mod 9)}$. Since $2^6 = 64 \equiv 1\text{ (mod 9)}$, multiplying by $2^6$ gives $2^{3+6n} \equiv -1\text{ (mod 9)}$.

The solutions that work are in the form $2^a+2^b$. Since $2^{3+6n}+1 \equiv 0\text{ (mod 9)}$, all of the solutions are in this form or this form multiplied by $2^x$ where $0 \leq 3+6n+x \leq 39$.

Now we just do casework: $$2^3+1: 2^0 \text{ to } 2^{36}$$ $$2^9+1: 2^0 \text{ to } 2^{30}$$ $$2^{15}+1: 2^0 \text{ to } 2^{24}$$ $$2^{21}+1: 2^0 \text{ to } 2^{18}$$ $$2^{27}+1: 2^0 \text{ to } 2^{12}$$ $$2^{33}+1: 2^0 \text{ to } 2^6$$ $$2^{39}+1: 2^0$$

So, the numerator is $37+31+25+19+13+7+1 = 133$. The denominator is just ${40 \choose 2}$, so the probability is $\frac{133}{780}$, and the answer is $133 + 780 = \boxed{913}$.

## Solution 3 (similar to 1)

As mentioned above, there are 780 possible combinations. Since we are essentially adding two powers of two together, thinking about the properties of this sum organizes our solution. All powers of two are even except for $2^0$, so we begin with labeling an entire group "where one of the 1s is in the rightmost spot". In an equation, $2^n+1$ is congruent to 0 mod 9. Instantly we think of $2^3=8$, and trial and error soon gives us $2^9=512$. Notice a pattern? Trying 2^15 = 32768 also works. You could go on, but basically all powers of two 3 mod 6 are 1 mod 9. (This is proven with the fact that $2^6$ is 1 mod 9) There are seven 3 mod 6 powers of 2 from 1-39, so for our first category there are seven "good" combinations.

Next, we look at powers of two exclusive from 1-39. Expressed in an equation, $2^n+2^m$ is congruent to 0 mod 9. Since nine is a relatively small number, we can treat powers of two as remainders of nine, to find "good" combinations quickly. Making a list of powers of two from 1-6 shows that the remainders when divided by nine are 2, 4, 8, 7, 5, and 1. This repeats itself every 6 powers. With simple counting we see there are seven powers of two that give remainders of 2, 4, and 8 each. Also there are six powers of two two that give remainders of 7, 5, and 1 each. Notice that each trio of powers correspond with another in the opposite trio. WLOG, if you choose two, (which there are seven to select) there are six different sevens to pair. This gives $7*6$. Repeat for the other two pairs.

In the end there are $7+42+42+42$ "good" combinations, for an answer of $133 + 780 = \boxed{913}$.

-jackshi2006

## See also

 2004 AIME II (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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