smc

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. $\boxed{142}$. 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.