AMC 8 · 2001 · #21
Grade 6 arithmeticnumber-theoryPick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The sum of all five numbers is locked at 75. So making the largest number as big as possible is the same as making the other four as small as possible — that is the Tool #11 (Work Backwards) move: start from the target ('largest as big as possible') and back into the smallest legal values for the other slots. Tool #2 (Make a Systematic List) lines up the five slots in increasing order so the median and 'distinct positive' constraints are easy to enforce one slot at a time.
The five numbers average to 15, so they add to 5 × 15 = 75 — a fixed total.
Mean × count = total. Holding the total fixed turns 'maximize one slot' into 'minimize the others'.
6.SP.B.5Work BackwardsSort the slots a₁ ＜ a₂ ＜ a₃ ＜ a₄ ＜ a₅; the median is the middle value, so a₃ = 18, with two numbers below and two above.
Sorting first makes the median rule visible: the third number is just the middle of the sorted list.
6.SP.A.3Make A Systematic ListThe two below-median slots must be distinct positive integers under 18, so take the smallest: a₁ = 1, a₂ = 2.
To shrink the two below-median slots as much as possible, pick the smallest two distinct positive integers.
6.EE.B.8Work Backwardsa₄ must be an integer above 18 and below a₅, so the smallest choice is a₄ = 19.
The slot right above the median has to clear 18, and the smallest integer that does that is 19.
6.EE.B.8Work BackwardsThe other four sum to 1 + 2 + 18 + 19 = 40, so a₅ = 75 - 40 = 35; order 1 ＜ 2 ＜ 18 ＜ 19 ＜ 35 holds → (D).
Once the other four are pinned to their minimums, the leftover from 75 is exactly the largest possible value.
6.EE.B.7Work BackwardsFixed mean means fixed total. To make one number as big as possible, push the other four down to their smallest legal values. Here the floors are 1, 2, 18, 19, leaving 75 - 40 = 35 for the largest — answer (D).