Problem 251

There are $1+2+..+199+200=\frac{(200)(201)}{2}=20100$ numbers in total. Let the median be $k$. We want to find the median $k$ such that
$$\frac{k(k+1)}{2}=20100/2,$$
or
$$k(k+1)=20100.$$
Note that $\sqrt{20100} \approx 142$. Plugging this value in as $k$ gives
$$\frac{1}{2}(142)(143)=10153.$$
$10153-142<10050$, so $142$ is the $152$nd and $153$rd numbers, and hence, our desired answer. $\fbox{(C) 142}$.
Note that we can derive $\sqrt{20100} \approx 142$ through the formula $$\sqrt{n} = \sqrt{a+b} \approx \sqrt{a} + \frac{b}{2\sqrt{a} + 1},$$
where $a$ is a perfect square less than or equal to $n$. We set $a$ to $19600$, so $\sqrt{a} = 140$, and $b = 500$. We then have $n \approx 140 + \frac{500}{2(140)+1} \approx 142$. ~approximation by ciceronii
Note by Fasolinka (use answer choices): Once you know that the answer is in the 140s range by the approximation, it is highly improbable for the answer to be anything but C.