AMC 10 · 2006 · #4
Grade 3 arithmeticPick an answer.
The display splits cleanly into an hour part and a minute part, so Tool #7 (Identify Subproblems) lets us maximize each part on its own and then add. Tool #14 (Extreme Principle) drives each part to its biggest digit sum — but the catch is that a bigger clock number is not always a bigger digit sum, so we test the boundary cases instead of just grabbing 12. Tool #3 (Eliminate Possibilities) guards against the trap answer that comes from wrongly picking the 12 o'clock hour.
Split the display into two parts
The two parts are independent, so each can be maximized alone.
The hour and the minutes are chosen independently, so the best full display is just the best hour paired with the best minute.
The hour and the minutes are chosen independently, so the best display is the best hour beside the best minute.
▸ Why?
Independent choices combine freely, so the best of each can be picked without spoiling the other.
▸ Why?
The display is its digits sitting in fixed places, so the digit sum only cares about each place separately.
Make the hour digits as big as possible
By digit sum the best hour is 9, not the largest hour.
A larger number can have smaller digits, so hunt for the biggest single digit, not the biggest hour.
1.NBT.B.2Extreme PrincipleMake the minute digits as big as possible
Each minute place has its own ceiling, giving 14.
Fill each place with the largest digit that place is allowed to hold, and the minutes reach 59.
1.NBT.B.2Extreme PrincipleAdd the two best parts
Adding the two best parts gives 23, choice (E).
The best hour and the best minute join into 9 : 59, the single display with the fattest digits.
2.NBT.B.5Eliminate PossibilitiesFatten each spot on its own — the hour 9 (not 12!) plus the minutes 59 give 9 + 5 + 9 = 23.
- Split the display into two parts
- Make the hour digits as big as possible
- Make the minute digits as big as possible
- Add the two best parts