Three distinct numbers are selected randomly from the set $\{1,2,3, \ldots, 40\}$. If the probability, that the selected numbers are in an increasing G.P., is $\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to __________ .

<p>Total number of ways of selecting $3$ distinct numbers from $\{1,2,\dots,40\}$ is</p> <p>$\binom{40}{3}=\frac{40\cdot 39\cdot 38}{6}=9880.$</p> <p>Now, three numbers in increasing G.P. can be written as</p> <p>$a,\; ar,\; ar^2 \quad (r>1).$</p> <p>Let $r=\dfrac{p}{q}$ in lowest form, where $p>q$ and $\gcd(p,q)=1$.</p> <p>Then</p> <p>$a,\; a\frac{p}{q},\; a\frac{p^2}{q^2}$</p> <p>must all be integers. So $a$ must be a multiple of $q^2$. Write</p> <p>$a=kq^2 \quad (k\in\mathbb{N}).$</p> <p>Then the triple becomes</p> <p>$kq^2,\; kpq,\; kp^2,$</p> <p>which is strictly increasing since $p>q$.</p> <p>Also all must be $\le 40$. Since</p> <p>$q^2 < pq < p^2,$</p> <p>it is enough to ensure</p> <p>$kp^2 \le 40 \;\Rightarrow\; k \le \left\lfloor \frac{40}{p^2}\right\rfloor.$</p> <p>So for each coprime pair $(p,q)$ with $p>q$, the number of valid triples equals $\left\lfloor \dfrac{40}{p^2}\right\rfloor$.</p> <p>Now $p^2\le 40 \Rightarrow p\le 6$. Count for each $p$:</p> <ul> <li><p>$p=2$: $q=1$ (1 value), $\left\lfloor \frac{40}{4}\right\rfloor=10$ $\Rightarrow 10$</p></li> <li><p>$p=3$: $q=1,2$ (2 values), $\left\lfloor \frac{40}{9}\right\rfloor=4$ $\Rightarrow 2\cdot 4=8$</p></li> <li><p>$p=4$: $q=1,3$ (2 values), $\left\lfloor \frac{40}{16}\right\rfloor=2$ $\Rightarrow 2\cdot 2=4$</p></li> <li><p>$p=5$: $q=1,2,3,4$ (4 values), $\left\lfloor \frac{40}{25}\right\rfloor=1$ $\Rightarrow 4\cdot 1=4$</p></li> <li><p>$p=6$: $q=1,5$ (2 values), $\left\lfloor \frac{40}{36}\right\rfloor=1$ $\Rightarrow 2\cdot 1=2$</p></li> </ul> <p>Hence number of favourable triples:</p> <p>$10+8+4+4+2=28.$</p> <p>Therefore required probability is</p> <p>$\frac{28}{9880}=\frac{7}{2470} \quad (\gcd(7,2470)=1).$</p> <p>So $m=7,\; n=2470 \Rightarrow m+n=2477.$</p>