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

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It's clear what the pairs <math>j, k</math> can look like.  If one is of the form <math>6n</math> (7 choices), the other must be of the form <math>6n + 3</math> (7 choices).  If one is of the form <math>6n + 1</math> (7 choices) the other must be of the form <math>6n + 4</math> (6 choices).  And if one is of the form <math>6n + 2</math> (7 choices), the other must be of the form <math>6n + 5</math> (6 choices).  This means that there are <math>7\cdot 7 + 7\cdot 6 + 7\cdot 6 = 49 + 42 +42 = 133</math> total "good" numbers.
 
It's clear what the pairs <math>j, k</math> can look like.  If one is of the form <math>6n</math> (7 choices), the other must be of the form <math>6n + 3</math> (7 choices).  If one is of the form <math>6n + 1</math> (7 choices) the other must be of the form <math>6n + 4</math> (6 choices).  And if one is of the form <math>6n + 2</math> (7 choices), the other must be of the form <math>6n + 5</math> (6 choices).  This means that there are <math>7\cdot 7 + 7\cdot 6 + 7\cdot 6 = 49 + 42 +42 = 133</math> total "good" numbers.
  
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So our [[probability]] is <math>\frac{133}{780}</math> and our answer is <math>133 + 780 = 913</math>.
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The [[probability]] is <math>\frac{133}{780}</math>, and the answer is <math>133 + 780 = 913</math>.
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== See also ==
 
== See also ==
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* [[2004 AIME II Problems/Problem 9 | Previous problem]]
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{{AIME box|year=2004|n=II|num-b=9|num-a=11}}
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* [[2004 AIME II Problems/Problem 11 | Next problem]]
 
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* [[2004 AIME II Problems]]
 
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 12:27, 19 April 2008

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.

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

See also

2004 AIME II (Problems • Answer Key • Resources)
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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