AMC 10 · 2019 · #3
Grade 3 arithmeticPick an answer.
Tool #16 (Change Focus): instead of asking 'how many draws guarantees 15 of one color', flip the question — 'what is the largest number of draws where we still avoid having 15 of any color?' Add one more to that worst case and the next ball MUST push some color to 15. Tool #2 (List): write out the maximum we can take of each color without hitting 15 — capped at 14 for the three big colors, and capped at the whole supply for the three small ones. Tool #3 (Eliminate): the choices 75, 76, 79, 84, 91 differ by only a few — once we compute the worst-case total 75, the answer is 75 + 1 = 76, picking (B).
Which colors can reach fifteen
Colors with few balls are impossible from the start.
Only colors with ≥ 15 balls can ever 'cross the line'.
3.OA.A.3Change Focus Count The ComplementBuild the worst case
Take fourteen of each possible color and all of the rest.
Pick the most balls you can without any single color reaching 15.
Pick the most balls you can without any single colour reaching the target, and that is the worst case.
▸ Why?
Once every colour is filled to just below the target, one more ball must push some colour over.
▸ Why?
Any smaller draw is already covered by this worst case, so nothing below it can be guaranteed.
Add up that case
Together that is 75 balls.
75 is the largest 'unlucky' draw — still no winning color.
2.NBT.B.5Make A Systematic ListDraw one more
One more draw settles it at 76.
One more draw past the worst case forces success.
1.OA.A.1Change Focus Count The ComplementRule out the other choices
Seventy-five fails and larger values are not minimal.
76 is the smallest that always works — the others are too small or too big.
1.NBT.B.3Eliminate PossibilitiesThis AMC 12 problem only needs Grade 3 "word-problem reasoning" you already know — count the worst-case 'unlucky' pulls (14 + 14 + 14 + 13 + 11 + 9 = 75), then add one more so a color is forced to 15: 76.