AMC 10 · 2018 · #10
Grade 6 arithmeticPick an answer.
Tool #14 (Extreme Principle): fewer distinct values means each value must carry more of the 2018 entries, so we push every non-mode value to the largest count it is allowed to have. The unique-mode rule sets that ceiling. Tool #4 (Introduce a Variable): name the number of non-mode values d and turn the words into a single inequality 9d ≥ 2008. Tool #11 (Work Backwards): start from the fixed total 2018, peel off the 10 slots the mode already uses, and see how many slots the other values must cover.
Cap every other value at nine
Every other value appears at most nine times.
A unique mode of 10 leaves 9 as the hard ceiling for everyone else, and using the fewest values means leaning on that ceiling.
6.SP.B.5Extreme PrincipleRemove the mode's ten slots
Remove the mode's ten slots.
Once the mode's share is set aside, only the leftover entries still need a home.
4.OA.A.3Work BackwardsTurn it into one inequality
The leftover becomes one inequality.
If each helper carries at most 9, then d helpers carry at most 9d, and that must reach the leftover total.
If each helper carries at most nine, then a fixed number of helpers can carry at most nine times that many.
▸ Why?
The leftover total is the helpers' values added together, so their sum has to reach it.
▸ Why?
Replacing every value by its own ceiling can only raise the total, so the ceiling really is a ceiling.
Divide and read the remainder
The remainder forces one more value.
A remainder of 1 means you can't quite finish with 223 groups of 9, so you must round up to one more value.
6.NS.B.2Extreme PrincipleAdd the mode back
Adding the mode back gives 225.
Count the helpers, then remember to count the mode too.
4.OA.A.3Work BackwardsA unique mode of 10 caps everyone else at 9, so cover the other 2008 entries in groups of 9 — that needs 224 values, plus the mode makes 225.
- Cap every other value at nine
- Remove the mode's ten slots
- Turn it into one inequality
- Divide and read the remainder
- Add the mode back