10.4 Let $n$ - be a natural number greater than 10. What digit can stand immediately after the decimal point in the decimal representation of the number $\sqrt{n^{2}+n}$? Provide all possible answers and prove that there are no others.
Solution. Method 1. $n^{2}+n=(n+0.5)^{2}-0.2510$ ), the digit immediately after the decimal point is no less than 4. That is, it is 4.

Method 2. The required digit is the last digit in the number $\left[10 \sqrt{n^{2}+n}\right]=$ $\left[\sqrt{100 n^{2}+100 n}\right]$. Note that $100 n^{2}+100 n100 n^{2}+80 n+16=(10 n+4)^{2}$. Then $\left[\sqrt{100 n^{2}+100 n}\right]=$ $10 n+4$, and the last digit of this number is 4.

Method 3. We have $n+0.4$ solved incorrectly, which may have led to an incorrect answer | 5 points |
| One of the two statements is proven: a) the required digit is less than 5; b) the required digit is greater than 3; plus an example is provided showing that the digit 4 is possible | 4 points |
| One of the two statements is proven: a) the required digit is less than 5; b) the required digit is greater than 3 | 3 points |
| The problem is correctly reduced to a system of inequalities or to an equation with an integer part | 2 points |
| The correct answer is illustrated with an example showing that the digit 4 is possible | 1 point |
| The correct answer without justification or an incorrect answer | 0 points |