Problem 20 · 2023 AMC 8
Hard
Arithmetic & Operations
sum-constraintcasework
Two integers are inserted into the list 3, 3, 8, 11, 28 to double its range. The mode and median remain unchanged. What is the maximum possible sum of two additional numbers?
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Answer: D — 60.
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Hint 1 of 2
‘Double the range’ just fixes how far apart the new min and max are (50 apart). To make the sum big, don't lower the min — keep it at 3 and shoot the max as high as allowed.
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Hint 2 of 2
Keeping min = 3 forces the new max to 3 + 50 = 53 (one insert, used up). Now the second insert should be as big as possible without breaking median = 8 or mode = 3 — so it has to stay below 8.
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Approach: push max up, then maximize the other insert under the median/mode constraints
- Doubling the range only sets the spread between min and max to 50 — it doesn't say which one moves. Since we want a large sum, keep the small end at 3 and stretch the top: the new max must be 3 + 50 = 53. That's one of the two inserts spent.
- Now place the second insert x as high as possible. The list has 7 numbers, so the median is the 4th when sorted: 3, 3, 8, 11, 28, 53 already sit there, and median must stay 8.
- If x were 8 or more, the 4th spot would no longer be 8 (and x = 8 would also create a second mode). So x ≤ 7 — and x = 7 keeps 3 the unique mode and 8 the median.
- Max sum = 53 + 7 = 60. Worth keeping: to maximize a sum under min/max/median/mode rules, fix the cheap constraints first (anchor the min, blow up the max), then push each remaining value to the edge of what the rules allow.
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