Rioplatense Mathematical Olympiad

Notice $N>34$ so that the first condition can hold.

First we will show that $N$ must be a prime power. Assume the contrary, so $N$ has two or more distinct prime factors. In this case, we can write $N=ab$ with $a,b$ relatively prime integers greater than 1. By the Chinese Remainder Theorem, there exists an integer $x$ such that $x\equiv 0(\mathrm{mod}a)$ and $x\equiv -1(\mathrm{mod}b)$. Thus $x$ and $x+1$ are not divisible by $N$, but their product is. Now, if we take 30 consecutive integers which include $x$ and $x+1$ but none of them is divisible by $N$ (this can be done because $N>30$), we see the second condition is not fulfilled.

So $N=p^k$, with $p$ prime and $k$ a positive integer. In fact, $k\ge 2$, because a product of integers is divisible by $p$ if and only if one of the factors is divisible by $p$. Hence if $k=1$ the first condition will not be fulfilled. Now let us find the possible values for $p$.

If $p\ge 37$, in any set of 34 consecutive integers there is at most one of them which is divisible by $p$. Therefore, the product of 34 consecutive integers can only be divisible by $N=p^k$ if one of those integers is divisible by $N$, because all factors $p$ must come from the same number. In conclusion, we need $p<37$ in order for the first condition to be plausible.

On the other hand, if $p\ge 29$, we consider the 30 consecutive integers $$p^{k-1},p^{k-1}+1,\dots ,p^{k-1}+29.$$ Since this list contains at least one more multiple of $p$ besides $p^k$, the product of these numbers is divisible by $N=p^k$. This contradicts the second desired condition unless one of these numbers is divisible by $N$. This can only happen if $p^k\le p^{k-1}+29$, or equivalently, $$p^{k-1}(p-1)\le 29.(\ast )$$ This rules out $p\ge 7$, as the left hand side of $(\ast )$ would be at least $7\cdot 6=42$ (recall that $k$ is at least 2). If $p=5$, for $N>34$ to hold, it must be $k\ge 3$. But in this case, the left hand side of $(\ast )$ is at least $5^2\cdot 4=100>29$. If $p=3$, it must be $k\ge 4$ and the left hand side of $(\ast )$ is at least $3^3\cdot 2=54>29$. Finally, if $p=2$, then $k\ge 6$ and the left hand side of $(\ast )$ is at least $2^5=32>29$. Having exhausted all cases, we conclude that the required conditions cannot hold if $p\ge 29$.

So we have $29<p<37$, and there is exactly one prime number in this range, which is 31. We will now prove that all numbers of the form $31^k$, where $k\ge 2$, are solutions. The second condition is satisfied, as we explained previously, because in any set of 30 consecutive integers there is at most one of them which is divisible by 31. For the first one, consider the 34 consecutive integers $31^{k-1},31^{k-1}+1,\dots ,31^{k-1}+33$. As before, we know that their product is divisible by $31^k$ because there is another multiple of 31 besides $31^{k-1}$ in the set (namely $31^{k-1}+31$). To prove that none of these numbers is divisible by $31^k$, it is enough to check that $31^{k-1}+33<31^k$. This is equivalent to $31^{k-1}(31-1)>33$, which is true for $k\ge 2$.

In conclusion, rioplatense numbers are those of the form $31^k$ for $k\ge 2$.