9.2. What digit can the number $f(x)=[x]+[3 x]+[6 x]$ end with, where $x$ is an arbitrary positive real number? Here $[x]$ denotes the integer part of the number $x$, that is, the greatest integer not exceeding $x$.


Answer: $0,1,3,4,6,7$.

Solution. Let $x=[x]+a$, where $0 \leq a<1$ is the fractional part of $x$. Then it is easy to understand that $f(x)=[x]+[3 x]+[6 x]=10[x]+[a]+[3 a]+[6 a]$, so the last digit does not depend on the integer part of $x$. Consider the possible values of its fractional part, dividing the interval $[0,1)$ into six equal intervals: $\left[0, \frac{1}{6}\right),\left[\frac{1}{6}, \frac{2}{6}\right), \ldots\left[\frac{4}{6}, \frac{5}{6}\right),\left[\frac{5}{6}, 1\right)$. The values of $[a]$ on all intervals are 0, the values of $[3a]$ are $0,0,1,1,2,2$ respectively, and the values of $[6a]$ are $0,1,2,3,4,5$. By adding the corresponding values for the same intervals, we get the set of possible last digits of the number $f(x)=[x]+[3 x]+[6 x]$: these will be $0,1,3,4,6,7$.

Grading criteria. Proven that the last digit of $x$ depends only on the fractional part of $x: 2$ points. Proven what the last digits of the numbers $[a],[3 a]$ and $[6 a]$ are depending on the interval: 4 points. Only examples are provided where the last digits are $0,1,3,4,6,7: 1$ point.