XXIX Rioplatense Mathematical Olympiad

We claim that, if the exponent of the greatest prime divisor $p$ of $n$ is greater than or equal to $3$, then $n$ is cuboso. In fact, at some step of the process, we will get $p^3$ on the board, after we divide by every smaller prime divisor and the remaining factors $p$.

Conversely, if a number is cuboso, then the exponent of its greatest prime divisor is at least $3$. In order to see that, first note that every number written on the list is always divisible by the greatest prime divisor of the first one. Also, at some step of the process we will get a perfect cube and, in particular, each of its prime divisors has an exponent greater than or equal to $3$.

It remains to count how many numbers $n<2022$ have its greatest prime divisor with an exponent at least $3$. We split the problem into cases according to the value of this prime $p$.

Case $p\ge 13$: We would have $n\ge 13^3=2197>2022$, which is not possible.

Case $p=11$: We have $n=11^{3}m=1331m$, the only solution is $m=1$.

Case $p=7$: We have $n=7^{3}m$ and so $m<6$. There are $5$ possibilities.

Case $p=5$: We have $n=5^{3}m$, so $m<17$ and $m$ must have only $2$, $3$ or $5$ as prime factors. Every integer between $1$ and $16$ except for $7$, $11$, $13$, $14$ works, so there are $12$ possibilities.

Case $p=3$: We have $n=3^{3}m$ and so $m<75$. The only possible prime factors of $m$ are $2$ or $3$. We split into cases according to the exponent $x$ of $3$ in the prime factorization of $m$.

If $x=3$, $m=27$, $54$ If $x=2$, $m=9$, $18$, $36$, $72$ If $x=1$, $m=3$, $6$, $12$, $24$, $48$ If $x=0$, $m=1$, $2$, $4$, $8$, $16$, $32$, $64$

So there are $2+4+5+7=18$ possibilities.

Case $p=2$: Here $n=2^{3}m$ and $m$ must have only $2$ as prime factor. The possibilities for $n$ are $8$: all the powers of $2$ from $2^3$ to $2^{10}$.

In conclusion, there exist $1+5+12+18+8=44$ cuboso numbers that are strictly less than $2022$.