AMC 8 · 2023 · #20

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

mean-median-mode-rangeoptimization-countingsystematic-enumeration caseworksystematic-enumerationbound-inequality-then-enumerate ↑ Prerequisites: mean-median-mode-rangemental-arithmetic

📏 Long solution 💡 4 insights

Problem

Insert two integers into the list 3, 3, 8, 11, 28 so that the new list of seven numbers has DOUBLE the original range while keeping the same mode and the same median. Among all valid pairs, what is the LARGEST possible sum of the two inserted numbers?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The new range of 50 can come from three different sources: the original 3 stays as the minimum and a new large y becomes the maximum, OR the original 28 stays as the maximum and a new small x becomes the minimum, OR BOTH endpoints are new numbers. Tool #2 (Systematic List) handles this cleanly: list the three cases, solve each one in isolation, then compare. Inside each case the median and mode constraints squeeze x and y into a narrow integer band, and Tool #6 (Guess & Check) picks the extreme integer in that band so the sum is as big as it can be. Tool #3 (Eliminate) finally checks the winning sum against the five answer choices so a typo cannot slip through.

1STEP 1

Sorted 3,3,8,11,28 has range 25, mode 3, median 8 — the trio we must preserve.

range=28-3=25, mode=3, median=8

2STEP 2

Median forces one insert below 8 and one above; mode bans 8,11,28 and ties — so x ＜ 8 ＜ y with y≠11,28.

x ≤ 8, y ≥ 8; x ≠ 8, y ≠ 8,11,28; x ≠ y → x ＜ 8 ＜ y

3STEP 3

Apply new range = 50 three ways: keep min 3, or keep max 28, or make both ends new — three cases.

Case 1: min=3, max=y; Case 2: min=x, max=28; Case 3: min=x, max=y

4STEP 4

Case 1: min stays 3, so y=53; x can be as big as 7 (below 8), giving 7+53=60.

y=53, x_max=7 → x+y=7+53=60

5STEP 5

Case 2: max stays 28, so x=-22; y at most 27, giving -22+27=5 — far below Case 1.

x=-22, y_max=27 → x+y=-22+27=5

6STEP 6

Case 3: both ends new, x ＜ 3 and y=x+50; biggest x is 2, so y=52 and 2+52=54.

x_max=2, y=x+50=52 → x+y=2+52=54

7STEP 7

Compare 60, 5, 54: Case 1 wins with 60, which is choice (D).

max(60, 5, 54)=60 → (D)

Answer

Sanity-check the winner {3,3,7,8,11,28,53}: range =53-3=50 (double of 25, correct), median is the 4th term =8 (unchanged), and 3 still appears twice while every other number appears once, so mode =3 (unchanged). The sum 7+53=60 matches choice (D). The other choices 56,57,58,61 would require either y ≠ 53 (breaking the range) or x ＞ 7 (which forces x=8 and ruins the mode/median by giving {3,3,8,8,11,28,53}, a bimodal list with median 8 but 3 no longer the unique mode). Everything lines up.

💡Key takeaway

This AMC 8 problem only needs Grade 6 statistics (range, mode, median) and a little inequality reasoning you already know!

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