<s>[INST] One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $163/4.$ Which integer was removed? 
One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $\dfrac{163}4$.  Which integer was removed?

Source. British Mathematical Olympiad 2010/11, Round 1, Problem 1
I was hoping if someone could spot the flaw in my working for this question.
Attempt.
I began by letting the integer that was removed be $x$.
Then: $$\frac{1 + 2 + \cdots + (x-1) + (x+1) +\cdots + n} {n-1} = \frac{163}{4}$$
There is two arithmetic sums in the denominator, the first from 1 to $x$ and the second from $x+1$ to $n$.
These are equal to $\frac{x(x-1)}{2}$ and $\frac{(n-x)(n+x+1)}{2}$, and subbing in to first equation this gives:
$$\frac{x(x-1) + (n-x)(n+x+1)}{2(n-1)} = \frac{163}{4}$$
which reduces to:
$$\frac{n^2 + n - 2x}{2(n-1)} = \frac {163}{4}$$
And then:
$$2(n^2 + n -2x) = 163(n-1)$$
At first I thought you could consider factors, as 163 was prime then:
$n-1 = 2$ giving $n = 3$ and $n^2 + n - 2x = 163$, which using $n=3$ gives $x= -75.5$ which isn't our positive integer.
I then tried considering a quadratic in $n$ and using the discriminant but again that just looked to give a negative value of $x.$ I would be grateful for any help
 [/INST] Say you're removing $x$ from the set $1, 2, \ldots, n$.  The average of the resulting numbers will be at least $n/2$ (if you remove $n$) and at most $(n+2)/2$ (if you remove $1$.)  So we have
$$ n/2 \le 163/4 \le (n+2)/2 $$
or, multiplying through by 4,
$$ 2n \le 163 \le 2n+2. $$.
So $n = 81$.  You don't have to explicitly work out the number being removed, but it's $(1 + 2 \ldots + 81) - (163/4) \times 80$ = $(81 \times 82)/2 - (163/4) \times 80 = 61$.
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