2020 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 11 Medium

What is the median of the following list of 4040 numbers?

1,2,3, \ldots, 2020,1^{2}, 2^{2}, 3^{2}, \ldots, 2020^{2}

  • A.

    1974.5

  • B.

    1975.5

  • C.

    1976.5

  • D.

    1977.5

  • E.

    1978.5

Answer:C

Solution 1: We can see that 44^{2} is less than 2020. Therefore, there are 1976 of the 4040 numbers after 2020 . Also, there are 2064 numbers that are under and equal to 2020 . Since 44^{2} is equal to 1936 , it, with the other squares, will shift our median's placement up 44 . We can find that the median of the whole set is 2020.5, and 2020.5-44 gives us 1976.5. Our answer (C) 1976.5

Solution 2: We want to know the 2020 th term and the 2021 th term to get the median. We know that 44^{2}=1936 So numbers 1^{2}, 2^{2}, \ldots, 44^{2} are in between 1 to 1936 . So the sum of 44 and 1936 will result in 1980 , which means that 1936 is the 1980 th number. Also, notice that 45^{2}=2025, which is larger than 2021 . Then the 2020 th term will be 1936+40=1976, and similarly the2021 th term will be 1977 . Solving for the median of the two numbers, we get (C) 1976.5

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