Problem 21 · 1988 AJHSME
Stretch
Algebra & Patterns
case-on-where-n-falls
A fifth number, n, is added to the set {3, 6, 9, 10} to make the mean of the set of five numbers equal to its median. The number of possible values of n is
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Answer: C — 3.
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Hint 1 of 2
The mean is easy — it's always (3 + 6 + 9 + 10 + n) ⁄ 5. The median is the tricky one: it's whichever number ends up in the middle once you sort, and that *depends on where n lands*. What are the possible middle numbers?
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Hint 2 of 2
Split into cases by where n falls: small (median stays 6), in the middle (median is n itself), or large (median stays 9). Set mean = median in each case — and then check the n you get actually belongs in that case.
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Approach: case on where n lands in the sorted list
- The mean is always (28 + n) ⁄ 5. The median is the 3rd value after sorting {3, 6, 9, 10, n}, which depends on n:
- • If n is small (n ≤ 6): the order is n, 3, 6, 9, 10 → median 6. Set (28 + n)⁄5 = 6 → n = 2, and 2 ≤ 6 ✓. • If n is in the middle (6 ≤ n ≤ 9): median is n. Set (28 + n)⁄5 = n → 28 + n = 5n → n = 7, and 6 ≤ 7 ≤ 9 ✓. • If n is large (n ≥ 9): median 9. Set (28 + n)⁄5 = 9 → n = 17, and 17 ≥ 9 ✓.
- All three candidates land inside their own case, so n = 2, 7, 17 all work — 3 values.
- Why this transfers: the median has no single formula — its value changes as the unknown crosses the other numbers. Whenever an unknown can sit in different positions of a sorted list, break into cases, solve each, and *verify the answer falls in the case you assumed* (a candidate that escapes its range is rejected).
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