Ana thinks about some different non-zero natural numbers, and for each of them, she writes on a card every positive divisor. Then, Ana hands all the cards to Maria, who groups them based on the number written on each, defining the sequence $(a_n)_{n \geq 1}$ for which $a_k$ is the number of cards with the number $k$, for any $k \geq 1$.
Finally, for each non-zero term $a_k$ of this sequence, Maria writes on the board the value of the number $k^{a_k}$. Show that, by looking at the board, we can determine at least one of the numbers Ana had in mind.
I have no idea how to approach this problem. I need help finding a solution. I have only observed that the product of Ana’s divisors is the same as the product of the numbers written on the board. I took the case with primes: Let's consider a prime number, say p, in Ana's set. The only divisors of a prime number are 1 and itself. Therefore, in Maria's sequence, there will be a term $a_p$ indicating the count of cards with the number p.
Now, Maria writes on the board the value $p^{a_p}$. Since p is a prime number, the only way to represent it in this form is if $a_p$ is 1. Thus, we can deduce the presence of the prime number p in Ana's set. But I don’t know how to generalize it when we have composite numbers.
Source: rmg 2023
Consider the prime factorization of all the positive integers written on the board.
All integers on the board are of form $p_1^{a_1}p_2^{a_2}\dots$ Write each integer on the board in the form $(p_1^{b_1}\cdot p_2^{b_2}\dots)^n$ where $\gcd(b_1,b_2,\dots)=1$ and label each $(p_1^{b_1}\cdot p_2^{b_2}\dots)^n = t_k^n$. Consider the set $S_{t_k}$ which consists of all powers of $t_k$ that are on the board.
Claim: $|S_{t_k}|$ gives the highest power of $t_k$ that divides some of the terms of Anna's sequence.
If $t_k^n$ is the highest power of $t_k$ dividing any term, then all powers of $t_k$ till $n$ act as divisors. Then our claim follows because while writing the terms down Maria has to write the first $n$ powers of $t_k$, each raised to some number depending on how many times they are repeated on the cards.
Suppose the largest number in Anna's sequence is $k$. It follows that out of all divisors of all numbers, $k$ will be their maximum (since the divisors of $n$ are $\le n$). Hence in order to find it we can just find the maximum value of $t_k^{|S_{t_k}|}$.
Claim: If the board has some integers of the form $n=p_1^{a_1}p_2^{a_2}\dots$ where we already have that $\gcd(a_1,a_2,\dots)=1$ then the largest multiple of $n$ is also on Anna's list.
Since this number can't be written in the form $a^b$ for integral $a>1$ and $b>2$, this integer is a divisor of exactly one integer $k$ (not necessarily the greatest) in Anna's list. That means all multiples of $n$ on the board are some divisors of $k$, and the largest such multiple of $n$ on the board is $k$ itself.
Example: suppose you see $$1, 49, 49, 32, 64, 8, 14, 28, 56, 27, 216, 81, 144, 324, 36, 43, 86$$ on the board. Since $14=7\cdot 2$, the greatest multiple of $14$, which is $56$ on the board, will be on Anna's mind. Considering all $t_k^{|S_{t_k}|}$, we get that out of $7^2$, $2^3$, $14$, $28$, $56$, $3^2$, $6^2$, $18$, $12$, $43$ and $86$, it is easy to tell $86$ is the maximum value that divides some of the terms of Anna's sequence, and is in turn the maximum value thereof.
Another example is that of $$1, 8, 27, 16, 216, 81, 144, 324, 36$$ wherefrom we get the maximum powers of all divisors $2^2$, $3^2$, $6^2$, $18$ and $12$, and the maximum divisor is $36$. Indeed the sequence was generated by using $12,18,36$.