Difference between revisions of "2004 AIME II Problems/Problem 10"

(Solution)
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Now we just do casework:
 
Now we just do casework:
<cmath>2^3+1: 2^0 \text{ to } 2^36</cmath>
+
<cmath>2^3+1: 2^0 \text{ to } 2^{36}</cmath>
<cmath>2^9+1: 2^0 \text{ to } 2^30</cmath>
+
<cmath>2^9+1: 2^0 \text{ to } 2^{30}</cmath>
<cmath>2^{15}+1: 2^0 \text{ to } 2^24</cmath>
+
<cmath>2^{15}+1: 2^0 \text{ to } 2^{24}</cmath>
<cmath>2^{21}+1: 2^0 \text{ to } 2^18</cmath>
+
<cmath>2^{21}+1: 2^0 \text{ to } 2^{18}</cmath>
<cmath>2^{27}+1: 2^0 \text{ to } 2^12</cmath>
+
<cmath>2^{27}+1: 2^0 \text{ to } 2^{12}</cmath>
 
<cmath>2^{33}+1: 2^0 \text{ to } 2^6</cmath>
 
<cmath>2^{33}+1: 2^0 \text{ to } 2^6</cmath>
 
<cmath>2^{39}+1: 2^0</cmath>
 
<cmath>2^{39}+1: 2^0</cmath>

Revision as of 19:11, 25 June 2016

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 $133 + 780 = \boxed{913}$.

See also

2004 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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