In equality
$$1 * 2 * 3 * 4 * 5 * ... * 60 * 61 * 62 = 2023$$Instead of each asterisk, you need to put one of the signs “+” (plus), “-” (minus), “•” (multiply) so that the equality becomes true. What is the smallest number of "•" characters that can be used?
The answer is $2$, achieved by putting a $\cdot$ between $1$ and $2$ and between $9$ and $10$ and a $+$ everywhere else. Obviously, we need at least one $\cdot$, since otherwise the result is at most $1+2+\dots+62<2023$. It remains to consider the case when there is exactly one $\cdot$.
Let $S=1+2+\dots+62$, which is odd. By putting a $\cdot$ between $a$ and $a+1$, the result will be $S-a-(a+1)+a(a+1)$, which is always even. Next, we can also flip some pluses to minuses, but this does not change the parity of the result. Hence, getting $2023$ with exactly one $\cdot$ is not possible.