AMC 10 · 2021 · #16

Grade 8 arithmetic
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Periodic Placement Formula
mean-median-mode-rangetriangular-numberssequences-arithmeticestimation identify-subproblemsguess-and-check ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 2 insights
Problem
One long increasing list is built by writing each integer from 1 to 200 exactly as many times as its own value: one 1, two 2s, three 3s, and so on up to two hundred 200s. Find the median of that whole list.

Pick an answer.

(A)
100.5
(B)
134
(C)
142
(D)
150.5
(E)
167
How to solve
Strategy Organize Information in More Ways

There are far too many entries to write out, so the list has to be re-organized into something countable. The useful reorganization is to stop tracking values and start tracking positions: because the list is already sorted, each value n owns one solid block of consecutive positions, and the only thing that matters is which block the middle position falls into. That converts the whole problem into running totals 1 + 2 + … + n, which have a pattern worth naming with a variable. Once the middle position is a specific number, finding the right block is a search over one unknown, and the near-square shape of the running total makes a square root an excellent first guess that two exact checks then confirm.

1STEP 1

Count the list's length

The total is a triangular number.

1 + 2 + … + 200 = (200 · 201)/2 = 20100
2STEP 2

Turn the median into a seat

The median averages the two middle seats.

median = (a₁₀₀₅₀ + a₁₀₀₅₁)/2, 10050 = 20100/2
3STEP 3

Give each block its range

Write each block's seat range.

T_n = (n(n+1))/2; block of n occupies seats T_n-1+1 through T_n
4STEP 4

Estimate with a square root

A square root estimates the block.

n²/2 ≈ 10050 → n ≈ √(20100), 141² = 19881 ＜ 20100 ＜ 20164 = 142²
5STEP 5

Check the candidate blocks

Both seats fall in the same block, so the answer is 142.

T₁₄₁ = (141 · 142)/2 = 10011 ＜ 10050, T₁₄₂ = (142 · 143)/2 = 10153 ≥ 10051 → median = (142 + 142)/2 = 142
Answer
142
The size is believable. The values 1 through 141 are 141 of the 200 possible values, yet they fill only 10011 of the 20100 seats — just under half — because the big values are repeated far more often. So the halfway seat should land just past 141, which it does. The repetition also explains why the median is much larger than 100.5, the middle of 1 through 200 when every value appears once; choice (A) is exactly that trap answer. The half-integer choices can be ruled out on structure alone: a median ending in .5 needs the two middle seats to hold different values, which only happens when seat 10050 is the very last seat of a block, that is when T_n = 10050 exactly. Since T₁₄₁ = 10011 and T₁₄₂ = 10153, no T_n equals 10050, so (A) and (D) are impossible. The remaining integers fail the seat test: T₁₃₄ = (134 · 135)/2 = 9045, so the 134 block ends well before seat 10050, and T₁₆₆ = (166 · 167)/2 = 13861, so the 167 block does not even begin until seat 13862, far past the middle. Only 142 survives.
💡Key takeaway

When values repeat different numbers of times, the median is decided by counting seats, not by looking at values: find the running total that first passes the halfway seat.

  • Count how long the list is
  • Turn the median into a position
  • Give each block its seat range
  • Estimate n with a square root
  • Check the two candidate blocks

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