The South African Mathematical Olympiad Third Round

All lower case variables in this solution denote integers. Let $X$ denote the set of visionary integers. We show that $X=\{\lfloor \frac{2020}{n}\rfloor :1\le n\le 2020\}$.

If $k$ is a visionary integer, then there exist $a>0$ and $b\ge 0$ such that $ak+b(k+1)=2020$, i.e., $2020=k(a+b)+b$, where $0\le b<a+b$. This implies that $k=\lfloor \frac{2020}{a+b}\rfloor$ and we also have $1\le a+b\le 2020$, since $a\ge 1$ and $k\ge 1$. Hence $k\in X$. Conversely, let $k\in X$, i.e., $k=\lfloor \frac{2020}{n}\rfloor$ for some $1\le n\le 2020$. Then $2020=kn+r$, where $0\le r<n$. Put $b=r$ and $a=n-r$. Then $a>0$ and $b\ge 0$, and we see that $ak+b(k+1)=2020$, i.e., $k$ is visionary.

In order to solve the problem, we need to find the cardinality of the set $X$. To this end, write $X$ as the disjoint union of $X_1=\{\lfloor \frac{2020}{n}\rfloor :1\le n<\sqrt{2020}\}$ and $X_2=\{\lfloor \frac{2020}{n}\rfloor :\sqrt{2020}<n\le 2020\}$. (Note that 2020 is not a square.) Also, since the smallest element of $X_1$ is $\lfloor \frac{2020}{\lfloor \sqrt{2020}\rfloor }\rfloor =\lfloor \frac{2020}{44}\rfloor =45$, and the largest element of $X_2$ is $\lfloor \frac{2020}{\lfloor \sqrt{2020}\rfloor +1}\rfloor =\lfloor \frac{2020}{45}\rfloor =44$, the sets $X_1$ and $X_2$ are indeed disjoint.

Let $1\le n_1<n_2\le \lfloor \sqrt{2020}\rfloor$. If $\lfloor \frac{2020}{n_1}\rfloor =\lfloor \frac{2020}{n_2}\rfloor$, then $0<\frac{2020}{n_1}-\frac{2020}{n_2}<1$, implying that $0<n_2-n_1<1$, an impossibility. This shows that $X_1$ has exactly $\lfloor \sqrt{2020}\rfloor =44$ elements.

Next, consider any $n$ such that $1\le n\le \lfloor \sqrt{2020}\rfloor$. We show that there exists a $q$, where $\sqrt{2020}<q\le 2020$, such that $\lfloor \frac{2020}{q}\rfloor =n$. By the Division Algorithm, there exist (unique) $q$ and $r$ such that $2020=qn+r$, where $0\le r<n$. Now if $q\le \lfloor \sqrt{2020}\rfloor$, then $2020=qn+r<qn+n=n(q+1)\le \lfloor \sqrt{2020}\rfloor \cdot (\lfloor \sqrt{2020}\rfloor +1)=44\cdot 45=1980$, a contradiction. So we have $\sqrt{2020}<q\le 2020$. Moreover, $n=\lfloor \frac{2020}{q}\rfloor$, as $2020=nq+r$, where $0\le r<n<q$. Finally, if $\sqrt{2020}<n\le 2020$, then $1\le \frac{2020}{n}<\sqrt{2020}$, so that also $1\le \lfloor \frac{2020}{n}\rfloor <\sqrt{2020}$, i.e., all elements of $X_2$ lie in the interval $[1,\lfloor \sqrt{2020}\rfloor ]$. This shows that $|X_2|=|X_1|$, and we conclude that $|X|=|X_1|+|X_2|=44+44=88$.