Problem 1945

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 $a k + b(k + 1) = 2020$, 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 $2020 = k n + r$, where $0 \le r < n$. Put $b = r$ and $a = n - r$. Then $a > 0$ and $b \ge 0$, and we see that $a k + b(k + 1) = 2020$, 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 $2020 = q n + r$, where $0 \le r < n$. Now if $q \le \lfloor \sqrt{2020} \rfloor$, then $2020 = q n + r < q n + n = n(q+1) \le \lfloor \sqrt{2020} \rfloor \cdot (\lfloor \sqrt{2020} \rfloor + 1) = 44 \cdot 45 = 1980$, a contradiction. So we have $\sqrt{2020} < q \le 2020$. Moreover, $n = \lfloor \frac{2020}{q} \rfloor$, as $2020 = n q + r$, 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 $|X| = |X_1| + |X_2| = 44 + 44 = 88$.