2. How many five-digit numbers are roots of the equation $x=[\sqrt{x}+1][\sqrt{x}]$? The symbol $[a]$ denotes the integer part of the number $a$, that is, the greatest integer not exceeding $a$.

(0. A. Pyayve)
Solution. Let $n=[\sqrt{x}]$, then $[\sqrt{x}+1]=[\sqrt{x}]+1=n+1$, which means $x=n(n+1)$.

All numbers of the form $x=n(n+1)$ are suitable, since for them $n<\sqrt{x}<n+1$, meaning $[\sqrt{x}]$ is indeed equal to $n$.

It remains to count the five-digit numbers of this form. Note that 99$\cdot$100 $<10000<100 \cdot 101$, $315 \cdot 316<100000<316 \cdot 317$, so $x$ is a five-digit number for $n$ from 100 to 315 inclusive.

Answer: 216 numbers.

Criteria. No less than 3 points if the participant is looking for numbers of the form $n(n+1)$; no less than 4 points if he found the number 315 as the "upper bound".

5 points - the answer differs from the correct one by 1 in either direction.

6 points - the correct solution and answer, but with a vague explanation: for example, it is not proven (or even mentioned) that all numbers of the form $n(n+1)$ indeed fit.