Problem 25 · 1986 AJHSME
Stretch
Number Theory
average-of-arithmetic-progression
Which of the following sets of whole numbers has the largest average?
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Answer: D — multiples of 5 between 1 and 101.
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Hint 1 of 3
You do NOT have to add up dozens of numbers. Each list is evenly spaced, so its values are perfectly balanced around their middle — the average is just the midpoint of the first and last terms.
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Hint 2 of 3
So for each set you only need two numbers: its smallest multiple and its largest multiple (up to 100). Average = (first + last) ⁄ 2.
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Hint 3 of 3
The winner will tend to be the one whose last term reaches closest to the top, 100.
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Approach: average of an evenly-spaced list = (first + last) ⁄ 2
- Each set is an evenly-spaced (arithmetic) list, and such lists are symmetric about their center — every term below the middle is matched by one equally far above. So the average equals the midpoint of the first and last terms; no long addition needed.
- Just grab first and last of each: 2's → (2 + 100)⁄2 = 51; 3's → (3 + 99)⁄2 = 51; 4's → (4 + 100)⁄2 = 52; 5's → (5 + 100)⁄2 = 52.5; 6's → (6 + 96)⁄2 = 51.
- The largest is 52.5, from the multiples of 5.
- Why 5 wins: its last multiple under 101 is 100 (hitting the ceiling) while its first term, 5, is small — so its first-and-last midpoint sits highest. Whenever you must average an evenly spaced list, reach for the (first + last)⁄2 shortcut.
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