Three distinct numbers are selected randomly from the set 1,2,3, , 40.

Question

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 \_\_\_\_\_\_\_\_\_\_ .

Answer: 2477

Accepted Answer

Total number of ways of selecting $3$ distinct numbers from ${1,2,\ldots ,40}$ is

$(\frac{40}{3})=\frac{40\cdot 39\cdot 38}{6}=9880.$

Now, three numbers in increasing G.P. can be written as

$a,\, ar,\, a{r}^{2}\, (r>1).$

Let $r=\frac{p}{q}$ in lowest form, where $p>q$ and $gcd(p,q)=1$.

Then

$a,\, a\frac{p}{q},\, a\frac{{p}^{2}}{{q}^{2}}$

must all be integers. So $a$ must be a multiple of ${q}^{2}$. Write

$a=k{q}^{2}\, (k\in N).$

Then the triple becomes

$k{q}^{2},\, kpq,\, k{p}^{2},$

which is strictly increasing since $p>q$.

Also all must be $\le 40$. Since

${q}^{2}it is enough to ensure

$k{p}\^{2}\le 40\, \Rightarrow \, k\le \lfloor \frac{40}{{p}\^{2}}\rfloor .$

So for each coprime pair $(p,q)$ with $p>q$, the number of valid triples equals $\lfloor \frac{40}{{p}\^{2}}\rfloor $.

Now ${p}\^{2}\le 40\Rightarrow p\le 6$. Count for each $p$:

$p=2$: $q=1$ (1 value), $\lfloor \frac{40}{4}\rfloor =10$ $\Rightarrow 10$

$p=3$: $q=1,2$ (2 values), $\lfloor \frac{40}{9}\rfloor =4$ $\Rightarrow 2\cdot 4=8$

$p=4$: $q=1,3$ (2 values), $\lfloor \frac{40}{16}\rfloor =2$ $\Rightarrow 2\cdot 2=4$

$p=5$: $q=1,2,3,4$ (4 values), $\lfloor \frac{40}{25}\rfloor =1$ $\Rightarrow 4\cdot 1=4$

$p=6$: $q=1,5$ (2 values), $\lfloor \frac{40}{36}\rfloor =1$ $\Rightarrow 2\cdot 1=2$

Hence number of favourable triples:

$10+8+4+4+2=28.$

Therefore required probability is

$\frac{28}{9880}=\frac{7}{2470}\, (gcd(7,2470)=1).$

So $

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