The average weight of $46$ weightlifters is $106$ kg. If none of the lifters weighs less than $82$ kg, at most how many of them can weigh more than $125$ kg?
We are given that $$\overline{X}=\dfrac{x_1+x_2+x_3+\ldots+x_{46}}{46}=106.$$ From here we can conclude $$x_1+x_2+x_3+\ldots+x_{46}=106\times46=4876 \text{ kg }$$ So the total weight of the weightlifters is $4876$ kg. We also know that $x_n\ge82$ kg for every $n$. I don't know what to do next. Thank you in advance!
On
Let divide the $n=46$ lifters into $h$ "heavy" lifters (weight more than 125) and $t=n-h$ "thin" lifters. Then the respective averages (denoted by $X$, $X_h$, $X_t$) are related by
$$ 106 = X = \frac{h}{n} X_h + \frac{t}{n} X_t =\frac{h}{n}(X_h - X_t) + X_t $$
Obviously $$ X_h > 125, $$ $$ 82 \le X_t \le 106 $$
and
$$ \frac{h}{n} = \frac{106 - X_t}{X_h - X_t} < \frac{106 - X_t}{125 - X_t} $$
The above is decreasing in $X_t \in [82,106]$, hence its maximum occurs at $X_t= 82$, so
$$ \frac{h}{46} < \frac{106-82}{125-46} \implies h < 25.674$$
and the maximum number of heavy lifters is $25$.
$ \newcommand{\ve}{\varepsilon} $ We can pair them off, one at a time. Say $x_1 = 82$ (the minimum weight), and then $x_2 = 125 + \varepsilon$, where $\ve$ is some small but positive number. The pair in total weigh $207 + \ve$, and there are $23$ such pairs. Thus, we have a total weight, at this point, of $4761 + 23 \ve$, but we also know to this to be equal to $4876$. Thus,
$$4761 + 23 \ve = 4876 \implies \ve = 5$$
So among them, there are $5$ kg extra per pair, and $23 \times 5 = 115$ in total spare. We can redistribute this to some pairs, with the lower-weight person receiving the weight. They each need $43 + \mu$ for some small positive $\mu$, and thus we can only give two people enough ($115/43 \approx 2.67$).
Thus, in total, $25$ people may exceed $125$ kg in weight (and $21$ do not).
To see that $26$ is not viable, suppose the remaining $20$ are at minimum weight. Then $26 \times 125 + 20 \times 82 = 4890$, just in excess of $4876$, with an average of $106.3 \ne 106$.