2012 AMC 10A Problems/Problem 25

Problem

Real numbers $x$ , $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ , $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$ ?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution

Without loss of generality, assume that $0 \le x \le y \le z \le n$ .

Then, the possible choices for $x$ , $y$ , and $z$ are represented by the expression $\frac{n^3}{3!}=\frac{n^3}{6}$ .

There are two restrictions: $x+1 \le y$ and $y+1 \le z$ .

Now, let $y'=y+1$ . Then, $x+2 \le y'$ and $y' \le z$ .

Then, let $x'=x+1$ . Combining the two inequalities gives us $x' \le y' \le z$ .

Since $0 \le x$ , then $2 \le x'$ . Thus, $2 \le x' \le y' \le z \le n$ .

There are $n-2$ ways to choose each number; the successful choices are represented by $\frac{(n-2)^3}{6}$ .

The probability then, is $\text{P}=\frac{\text{successful}}{\text{possible}}=\frac{\frac{(n-2)^3}{6}}{\frac{n^3}{6}}$ which must be greater than $\frac{1}{2}$ .

Plug in values. Trying $(\text{C})$ gives us $\frac{343}{729}$ , which is less than $\frac{1}{2}$ . Try the next integer, $10$ , which gives us $\frac{512}{1000}$ which is greater than $\frac{1}{2}$ .

Thus, our answer is $(\text{D})$ .

2012 AMC 10A (Problems • Answer Key • Resources)
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2012 AMC 10A Problems/Problem 25

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