Problem 19 · 1992 AJHSME
Hard
Algebra & Patterns
extremal
The distance between the 5th and 26th exits on an interstate highway is 118 miles. If any two exits are at least 5 miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the 5th and 26th exits?
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Answer: C — 18 miles.
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Hint 1 of 3
You have a fixed 118 miles to split among the gaps. To make ONE gap as long as possible, what should you do with all the OTHER gaps?
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Hint 2 of 3
Maximize-one-thing problems: push everything else to its limit in the opposite direction. Squeeze every other gap down to its smallest allowed size, freeing the leftover for your one big gap.
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Hint 3 of 3
Count the gaps carefully — from exit 5 to exit 26 there are 26 − 5 = 21 gaps, not 21 exits. The minimum each gap can be is 5 miles.
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Approach: shrink every other gap to the 5-mile minimum, leaving the rest for one
- From the 5th to the 26th exit there are 26 − 5 = 21 gaps sharing the 118 miles. To stretch one gap as far as possible, make the other 20 as short as the rules allow: 5 miles each, using 20 × 5 = 100 miles.
- Whatever's left goes into the one big gap: 118 − 100 = 18 miles.
- Why this transfers: to maximize one quantity under a fixed total, minimize all the others to their limits — that frees up the most for the one you care about. (To minimize one instead, you'd maximize the others.)
- Sanity check: 18 + 100 = 118 exactly, and 18 comfortably beats the 5-mile minimum, so this arrangement is legal.
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