AMC 10 · 2020 · #11

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

mean-median-mode-rangeperfect-squaressystematic-enumerationfloor-function easier-related-problemidentify-subproblems ↑ Prerequisites: mean-median-mode-rangeperfect-squares

📏 Long solution 💡 3 insights

Problem

Sort the 4040-number list {1, 2, 3, …, 2020, 1², 2², 3², …, 2020²} in non-decreasing order, then find the median (the average of the 2020-th and 2021-st terms).

Pick an answer.

(A)

1974.5

(B)

1975.5

(C)

1976.5

(D)

1977.5

(E)

1978.5

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

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.

1STEP 1

Warm up on N=4: sorted it's 1,1,2,3,4,4,9,16, so the median is 3.5 — and it lands among the small integers, not the big squares.

sorted: 1, 1, 2, 3, 3, 4, 4, 9, 16 → median = 3.5

2STEP 2

Back to N=2020: since 44 × 44 = 1936 ≤ 2020 ＜ 2025 = 45 × 45, the squares under 2020 are exactly 1², 2², …, 44² — just 44 of them.

44 × 44 = 1936 ≤ 2020 ＜ 2025 = 45 × 45

3STEP 3

For any integer k in the band 1936 ≤ k ≤ 2024, exactly 44 squares sit at or below it, so its rank is k + 44.

rank(k) = k + 44 for 1936 ≤ k ≤ 2024

4STEP 4

The 2020-th term: solve k + 44 = 2020 to get k = 1976, and 1976 really is in the integer list.

rank(1976) = 1976 + 44 = 2020 → S₂020 = 1976

5STEP 5

The 2021-st term: the next value up is the integer 1977, still below the next square 45² = 2025.

S₂021 = 1977 (since 1977 ＜ 2025 = 45²)

6STEP 6

Average the two middle terms: $\frac{1976 + 1977}{2}$ = 1976.5.

median =

\frac{1976 + 1977}{2}

= 1976.5

7STEP 7

Only choice (C) equals 1976.5; the others come from miscounting the squares by one or two.

1976.5 → (C)

Answer

1976.5

All five answer choices cluster around 1976.5, hinting the median should be near the integer whose rank shifts by about 44 (the count of squares ≤ 2020). Our value 1976.5 sits exactly 43.5 below 2020, matching the 44-square shift up to the half-step from averaging two consecutive integers. Magnitude and sign are both reasonable.

💡Key takeaway

This AMC 10 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).

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