Difference between revisions of "2020 AMC 10A Problems/Problem 11"
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As we are trying to find the median of a <math>4040</math>-term set, we must find the average of the <math>2020</math>th and <math>2021</math>st terms. <br> | As we are trying to find the median of a <math>4040</math>-term set, we must find the average of the <math>2020</math>th and <math>2021</math>st terms. <br> | ||
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− | Since <math>45^2 = 2025</math> is slightly greater than <math>2020</math>, we know that the <math>44</math> perfect squares <math>1^2</math> through <math>44^2</math> are less than <math>2020</math>, and the rest are greater. Thus, from the number <math>1</math> to the number <math>2020</math>, there are <math>2020 + 44 = 2064</math> terms. Since <math>44^2</math> is <math>44 + 45 = 89</math> less than <math>45^2 = 2025</math> and <math>84</math> less than <math>2020</math>, we will only need to consider the perfect square terms going down from the <math>2064</math>th term, <math>2020</math>, after going down <math>84</math> terms. Since the <math>2020</math>th and <math>2021</math>st terms are only <math>44</math> and <math>43</math> terms away from the <math>2064</math>th term, we can simply subtract <math>44</math> from <math>2020</math> and <math>43</math> from <math>2020</math> to get the two terms, which are <math>1976</math> and <math>1977</math>. Averaging the two, we get <math>\boxed{\textbf{(C) } 1976.5}.</math> | + | Since <math>45^2 = 2025</math> is slightly greater than <math>2020</math>, we know that the <math>44</math> perfect squares <math>1^2</math> through <math>44^2</math> are less than <math>2020</math>, and the rest are greater. Thus, from the number <math>1</math> to the number <math>2020</math>, there are <math>2020 + 44 = 2064</math> terms. Since <math>44^2</math> is <math>44 + 45 = 89</math> less than <math>45^2 = 2025</math> and <math>84</math> less than <math>2020</math>, we will only need to consider the perfect square terms going down from the <math>2064</math>th term, <math>2020</math>, after going down <math>84</math> terms. Since the <math>2020</math>th and <math>2021</math>st terms are only <math>44</math> and <math>43</math> terms away from the <math>2064</math>th term, we can simply subtract <math>44</math> from <math>2020</math> and <math>43</math> from <math>2020</math> to get the two terms, which are <math>1976</math> and <math>1977</math>. Averaging the two, we get <math>\boxed{\textbf{(C) } 1976.5}.</math> ~[[User:emerald_block|emerald_block]] |
==Video Solution== | ==Video Solution== |
Revision as of 17:25, 1 February 2020
- The following problem is from both the 2020 AMC 12A #8 and 2020 AMC 10A #11, so both problems redirect to this page.
Problem 11
What is the median of the following list of numbers
Solution 1
We can see that is less than 2020. Therefore, there are of the numbers after . Also, there are numbers that are under and equal to . Since is it, with the other squares will shift our median's placement up . We can find that the median of the whole set is , and gives us . Our answer is .
~aryam
Solution 2
As we are trying to find the median of a -term set, we must find the average of the th and st terms.
Since is slightly greater than , we know that the perfect squares through are less than , and the rest are greater. Thus, from the number to the number , there are terms. Since is less than and less than , we will only need to consider the perfect square terms going down from the th term, , after going down terms. Since the th and st terms are only and terms away from the th term, we can simply subtract from and from to get the two terms, which are and . Averaging the two, we get ~emerald_block
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 10 Problems and Solutions |
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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.