## Task A-1.2.

All natural numbers from 1 to some number were written on a board. After one of the numbers was erased, the arithmetic mean of the remaining numbers on the board is $\frac{673}{18}$. Which number was erased?
## First Solution.

Let the largest written natural number be $n$, and the erased number be $k$. The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

The condition of the problem can be written as

$$
\frac{\frac{n(n+1)}{2}-k}{n-1}=\frac{673}{18}
$$

If the erased number $k$ were $n$, we would get the smallest possible arithmetic mean. It would be

$$
\frac{\frac{n(n+1)}{2}-n}{n-1}=\frac{n}{2}
$$

and it is less than or equal to the obtained arithmetic mean:

$$
\frac{n}{2} \leqslant \frac{673}{18}
$$

If the erased number $k$ were 1, we would get the largest possible arithmetic mean. It would be

$$
\frac{\frac{n(n+1)}{2}-1}{n-1}=\frac{n+1}{2}
$$

and it is greater than or equal to the obtained arithmetic mean:

$$
\frac{n+1}{2} \geqslant \frac{673}{18}
$$

From the obtained inequalities, it follows that $72<n<75$, so $n=73$ or $n=74$.

If $n=73$, then

$$
\frac{673}{18}=\frac{\frac{73 \cdot(73+1)}{2}-k}{73-1}
$$

From this, we get $k=9$.

If $n=74$, we get

$$
\frac{673}{18}=\frac{\frac{74 \cdot(74+1)}{2}-k}{74-1}
$$

Solving this equation, we get $k=\frac{821}{18}$, which is not a natural number, so we conclude that $n$ cannot be 74.

We conclude that the number erased from the board is 9.