4. The positive integers from 1 to $\mathrm{n}$ inclusive are written on a whiteboard. After one number is erased, the average (arithmetic mean) of the remaining $n-1$ numbers is 22 . Knowing that $\mathrm{n}$ is odd, determine $\mathrm{n}$ and the number that was erased. Explain your reasoning.
Soln.
When we erase one number from $1,2, \ldots, \mathrm{n}$, the minimum sum of the remaining numbers is $\frac{(n-1) n}{2}$, and the maximum sum is: $\frac{n(n+1)}{2}-1=\frac{(n+2)(n-1)}{2}$
[1 mk]

Hence the average of the remaining numbers lies between $\frac{n}{2}$ and $\frac{(n+2)}{2}$. This gives $\frac{n}{2} \leq 22 \leq \frac{(n+2)}{2}$, whence $\mathrm{n}=42,43$ or 44 .
Thus $\mathrm{n}=43$. $\quad[1 \mathrm{mk}]$
Now $42 \times 22=924$ and the sum of all numbers from 1 to 43 is 946 . So the number erased is 22 . [1 mk]