Competition · AMC preparation · step 4 of 4
AMC 8 · 2005 · #16
Grade 4 countingPick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
"Be certain" is the tell for Tool #14 (Extreme Principle): the answer is whatever the worst-case sequence forces, not what a lucky pull might give. The worst case is the one that postpones reaching 5 of any one color as long as possible — spread the pulls as evenly across the three colors as we can. Tool #9 (Solve an Easier Problem) makes the idea concrete by warming up on a smaller version (say, 3 of one color from 2 colors) so the pattern is visible. Tool #2 (Systematic List) records the worst-case tally color by color so the count is hard to miscount.
Try an easier version
Warm up small: 2 colors, want 3 of one. Worst luck gives 2 of each, then the next pull forces a third — pattern (target−1)×(colors)+1.
Grade 3 sees multiplication as "equal groups." Here the groups are colors, and each group can hold up to target - 1 socks before the rule kicks in.
3.OA.A.1Solve An Easier Related ProblemBuild the worst case
Now the real case: 3 colors, target 5. Worst luck is 4 of each color before any hits 5 — tally red 4, white 4, blue 4 = 12.
3 groups of 4 is the Grade 3 "equal groups" picture: 3 × 4 = 12. None of the three colors has reached 5 yet.
3.OA.A.1Make A Systematic ListAdd one more sock
Extreme principle: after 12 socks (4 each), the next pull is red, white, or blue and lifts that color to 5 — so 13 guarantees it, 12 can't.
Worst case plus one is the Grade 3 two-step move: compute the unluckiest total, then add 1 to push it past the threshold.
Pulling thirteen socks forces five of them to share one color.
▸ Why?
If no single color reached five, every color would have to stop at four or fewer, and three colors each capped at four hold at most twelve socks together.
▸ Why?
Each sock wears exactly one of the three colors with none left over, so the socks pulled are just the red, white, and blue tallies added back up.
▸ Why?
Three colors, each holding at most four, make three equal groups of four, which is twelve.
▸ Why?
Thirteen socks are more than the twelve places a four-per-color cap allows, so they cannot each take a separate under-the-cap place — at least one color is forced to hold a fifth sock.
Check that 12 fails
Check nothing smaller works: with only 12 pulled, the split 4+4+4 leaves every color at 4 — none at 5 — so 12 is no guarantee.
Showing a single bad scenario is enough to knock out a candidate answer — the Grade 4 "check whether the answer makes sense" habit, used in reverse.
4.OA.A.3Extreme Principle"Be certain" problems are solved by the unluckiest run. Spread your pulls as evenly as possible across the categories, then add one more — that is the guarantee.
- Try an easier version
- Build the worst case
- Add one more sock
- Check that 12 fails
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