Difference between revisions of "2021 AMC 12A Problems/Problem 16"

Revision as of 14:37, 11 February 2021

Problem

In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ . $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \cdot, 200, 200, \cdot , 200$ What is the median of the numbers in this list?

Solution

Solution 1

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

See problem 1.

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 3: / Line 3: @@
 ==Solution==
-The solutions will be posted once the problems are posted.
+===Solution 1===
+There are <math>1+2+..+199+200=\frac{(200)(201)}{2}=20100</math> numbers in total. Let the median be <math>k</math>. We want to find the median <math>k</math> such that
+<cmath> \frac{k(k+1)}{2}=20100/2,</cmath>
+or
+<cmath> k(k+1)=20100.</cmath>
+Note that <math>\sqrt{20100} \approx 142</math>. Plugging this value in as <math>k</math> gives
+<cmath>\frac{1}{2}(142)(143)=10153.</cmath>
+<math>10153-142<10050</math>, so <math>142</math> is the <math>152</math>nd and <math>153</math>rd numbers, and hence, our desired answer. <math>\fbox{(C) 142}.</math>.
 ==Note==
 See [[2021 AMC 12A Problems/Problem 1|problem 1]].

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