AMC 10 · 2020 · #8
Grade 6 arithmeticPick an answer.
The list is huge (4040 entries), so Tool #9 (Easier Problem): first solve the same setup with N=4 in place of 2020 and watch what happens. From the small case we see that for each integer k, the squares 1², 2², …, k_max² that are ≤ k slip in among the small numbers and "push" the median value down. Tool #2 (Systematic List): count exactly how many list entries are ≤ k for each candidate k near the middle. Tool #5 (Pattern): the rank of k equals k + (number of squares ≤ k). Tool #3 (Eliminate): match the final median to the five choices.
Try a small case
A small case shows how they interleave.
Grade 6 median: shrink the problem to a list you can sort by hand.
6.SP.A.3Solve An Easier Related ProblemCount the small squares
Count how many squares stay below 2020.
Grade 5 multi-digit multiplication: 44 × 44 and 45 × 45 pin down the squares that fit.
5.NBT.B.5Make A Systematic ListWrite the rank formula
Each rank is shifted by that count.
Grade 4 pattern rule: each integer in that band picks up the same 44 extra entries from squares.
Each integer in the band picks up the same number of extra entries from the squares.
▸ Why?
The extra entries arrive at the same fixed rate across the band, so the ranks climb by a constant step.
▸ Why?
The median is whatever sits in the middle slot of the sorted list, so only the rank matters.
Find the 2020th value
Invert the formula to find it.
Grade 4 word-problem subtraction: 2020 - 44 = 1976 undoes the rank shift.
4.OA.A.3Make A Systematic ListFind the 2021st value
The next one works the same way.
Grade 4 comparing whole numbers: 1977 ＜ 2025, so 1977 comes first.
4.NBT.A.2Make A Systematic ListAverage them
Take their average.
Grade 6 median: average the two middle entries of an even-length sorted list.
6.SP.A.3Look For A PatternMatch the choice
The median is 1976.5.
Grade 4 comparison: only one choice equals 1976.5.
4.NBT.A.2Eliminate PossibilitiesThis AMC 12 problem only needs Grade 6 median you already know! The trick: the squares 1, 4, 9, 16, … that stay below 2020 are exactly 1² through 44² (since 44² = 1936 but 45² = 2025 is too big). So between 1936 and 2024, every integer k has rank k + 44. Solving k + 44 = 2020 gives k = 1976, the 2020-th entry. The 2021-st is 1977. Median = (1976 + 1977)/2 = 1976.5, answer (C).