The Median

We can graph $y=\frac{m}{x}$ , $x=10$ , and $y=10$ , as shown below.

Note that the red line represents $y=\frac{m}{x}$ when $m=50$ .

$P(xy<m)=1-P(xy>m)$ . The upper right area formed by the blue, red, and green lines represents $100P(xy>m)$ and is equal to

$\int_{\frac{m}{10}}^{10} (10-\frac{m}{x})dx=[10x-m\ln(x)]_{\frac{m}{10}}^{10}=100-m\ln(10)-m+m\ln(\frac{m}{10})$

$=100-m-m(\ln(10)-\ln(\frac{m}{10}))=100-m(1+\ln(\frac{100}{m}))$

Therefore, $P(xy<m)=1-P(xy>m)=1-(1-\frac{m(1+\ln(\frac{100}{m}))}{100})=\frac{m(1+\ln(\frac{100}{m}))}{100}$

Solving the equation $\frac{m(1+\ln(\frac{100}{m}))}{100}=0.5$ graphically, we get $m=18.668$ , so the answer is $\boxed{18668}$ .

Let us choose an real number $0≤y≤10$ . The probability that the y is in between $y$ and $y+dy$ is $\frac{dy}{10}$ . Now, for $y≤\frac{m}{10}$ , $x$ can be anything in the interval $[0,10]$ . So, in this case the probability of choosing $x$ obeying the given condition is exactly $1$ .

So, the probability is $P_1=1×\frac{m}{10}\frac{1}{10}=\frac{m}{100}$ .

Now if $y≥\frac{m}{10}$ , then $x$ has to be chosen from the interval $[0,\frac{m}{y}]$ ,and the probability of choosing that $x$ is $\frac{m}{10y}$ .

Then the total probability is
$P_2=\frac{m}{100}\int_{\frac{m}{10}}^{10} \frac{dy}{y}=\frac{m}{100}ln(\frac{100}{m})$ .

So, $0.5=P=P_1+P_2=\frac{m}{100}(1+(ln\frac{100}{m})$ .

Solving m we get $m=0.18668..$ . So, $[1000m]=18668$ .