Erased number? A set of consecutive positive integers starting with 1 is written on the board. A student came along and erased one number. Average of remaining numbers is 61 15/20 . What was the number erased
 A: Let $n$ be the last number written. Lets say that $m$ is the erased number.
Then the sum of the numbers on the board is $\frac{n(n+1)}{2}-m$. Their average then is
$$\frac{\frac{n(n+1)}{2}-m}{n-1}=61 \frac{15}{20}$$
Multiplying by 2 you get
$$\frac{n(n+1)-2m}{n-1}=122\frac{3}{2}$$
$$\frac{n^2+n-2}{n-1}+\frac{2}{n-1}-\frac{2m}{n-1}=123\frac{1}{2}$$
$$n+2+\frac{2-2m}{n-1}=123 \frac{1}{2}.\tag{$*$}$$
Now, since $1 \leq m \leq n$ we have
$$-2 \leq \frac{2-2m}{n-1} \leq 0 \,.$$
Using the fact that $n+2$ is an integer and $-2 \leq \frac{2-2m}{n-1} \leq 0 \,,$ in $(*)$, you see immediately that there are only two possibilities:
Case 1:
$n+2=124$ and $\frac{2-2m}{n-1}=-\frac{1}{2}$
Case 2:
$n+2=125$ and $\frac{2-2m}{n-1}=-\frac{3}{2}$
A: The average is $61 \frac 34$, and erasing one number can only move it by $\frac 12^+$, so the maximum had to be $122$ or $123$  But the denominator of the average is $4$, so the number of entries must be a multiple of $4$.  This is a contradiction and there is no solution.
Added:  There are $124$.  The average starts out $62 \frac 12$, then the $3$ of $103$ was erased, changing it to $10.$  This reduces the sum by $93$ and the average by $\frac 34.$  Some will think this unfair, some will think it clever.
