British Maths Olympiad 2010 Round 1 Problem 1

My solution to an Olympiad problem.

By Vamshi Jandhyala in mathematics

September 13, 2019

One number is removed from the set of integers from $1$ to $n$. The average of the remaining numbers is $40\frac{3}{4}$. Which integer was removed?

Solution

Let the integer removed be $x$. The average of the remaining numbers is

$$ \begin{align} \frac{n(n+1)}{2(n-1)}-\frac{x}{n-1} &&=&& 40\frac{3}{4} \
\implies \frac{(n-1)^{2}+3(n-1)+2}{2(n-1)}-\frac{x}{n-1} &&=&& 40\frac{3}{4} \
\implies \frac{n-1}{2} + \frac{3}{2} + \frac{1}{n-1} - \frac{x}{n-1} &&=&& 40\frac{3}{4} \
\implies \frac{n-1}{2} = 40 \land \frac{x-1}{n-1} = \frac{3}{4} \end{align} $$

From the above we get $n=81$ and $x=61$.