AMC 10 · 2006 · #4
Grade 3 arithmeticA digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A digital watch shows the time as an hour and a minute, in $12$-hour form with AM or PM. Among every time it can show, find the largest possible sum of the digits on the display.
Givens: The watch shows hours and minutes in $12$-hour format (with AM/PM), for example $9\!:\!59$; The hour part can be any whole number from $1$ to $12$; The minute part can be any whole number from $00$ to $59$, written with two digits; Answer choices: (A) $17$, (B) $19$, (C) $21$, (D) $22$, (E) $23$
Unknowns: The greatest possible sum of all the digits shown at one time
Understand
Restated: A digital watch shows the time as an hour and a minute, in $12$-hour form with AM or PM. Among every time it can show, find the largest possible sum of the digits on the display.
Givens: The watch shows hours and minutes in $12$-hour format (with AM/PM), for example $9\!:\!59$; The hour part can be any whole number from $1$ to $12$; The minute part can be any whole number from $00$ to $59$, written with two digits; Answer choices: (A) $17$, (B) $19$, (C) $21$, (D) $22$, (E) $23$
Plan
Primary tool: #14 Extreme Principle
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
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.
Execute — Answer: E
3.MD.A.1 Step 1 Split the display into two parts
- The digits on the watch come from two separate boxes: the hour (a number from $1$ to $12$) and the minutes (a two-digit number from $00$ to $59$).
- The AM/PM and the colon carry no digits.
- Since the hour choice and the minute choice do not affect each other, we can make each part as large as possible on its own and then add the two best sums together.
💡 The hour and the minutes are chosen independently, so the best full display is just the best hour paired with the best minute.
1.NBT.B.2 Step 2 Make the hour digits as big as possible
- The hour runs $1$ through $12$.
- It is tempting to grab $12$ because it is the largest number, but its digit sum is only $1 + 2 = 3$.
- A single-digit hour keeps everything in one place, and the largest single digit is $9$, giving a digit sum of $9$.
- Checking the two-digit hours $10, 11, 12$ gives sums $1, 2, 3$ — all far smaller.
- So the hour $9$ wins with a digit sum of $9$.
💡 A larger number can have smaller digits, so hunt for the biggest single digit, not the biggest hour.
1.NBT.B.2 Step 3 Make the minute digits as big as possible
- The minutes are two digits.
- The tens digit can only be $0,1,2,3,4,$ or $5$ (minutes never reach $60$), so its largest value is $5$.
- The ones digit is free to be any digit $0$ through $9$, so its largest value is $9$.
- Putting the biggest allowed digit in each place gives $59$, with digit sum $5 + 9 = 14$.
💡 Fill each place with the largest digit that place is allowed to hold, and the minutes reach $59$.
2.NBT.B.5 Step 4 Add the two best parts
- The best hour contributes $9$ and the best minutes contribute $14$, so the display $9\!:\!59$ gives a total digit sum of $9 + 14 = 23$.
- That matches choice (E).
- The trap answer $17$ comes from wrongly using the hour $12$: $1 + 2 + 5 + 9 = 17$, which is choice (A) — a good reminder that $9$ beats $12$ for digit sums.
💡 The best hour and the best minute join into $9\!:\!59$, the single display with the fattest digits.
3.MD.A.1 The digits on the watch come from two separate boxes: the hour (a number from $1 1.NBT.B.2 The hour runs $1$ through $12$. It is tempting to grab $12$ because it is the la 1.NBT.B.2 The minutes are two digits. The tens digit can only be $0,1,2,3,4,$ or $5$ (minu 2.NBT.B.5 The best hour contributes $9$ and the best minutes contribute $14$, so the displ Review
Reasonableness: The display $9\!:\!59$ uses digits $9, 5, 9$, and $9 + 5 + 9 = 23$, matching (E). No single position can do better: the ones digit is already maxed at $9$, the minute tens digit is capped at $5$, and no hour beats a lone $9$ (since $10,11,12$ give digit sums $1,2,3$). The biggest possible sum is therefore $9 + 5 + 9 = 23$, and it cannot be pushed higher, so choices $17$ through $22$ are all beaten.
Alternative: Bound each digit slot separately and add the ceilings: the largest useful hour digit sum is $9$ (from the hour $9$), the minute tens digit maxes at $5$, and the minute ones digit maxes at $9$. Adding the caps gives $9 + 5 + 9 = 23$, and because the time $9\!:\!59$ actually hits all three caps at once, $23$ is truly reachable — confirming (E).
CCSS standards used (min grade 3)
3.MD.A.1Tell and write time to the nearest minute (Knowing the display's hours run $1$–$12$ and its minutes run $00$–$59$, which fixes the allowed digits.)1.NBT.B.2Understand that the two digits of a two-digit number represent amounts of tens and ones (Seeing that the minute tens digit tops out at $5$ while the ones digit can reach $9$, and that the hour $9$ beats $12$ for digit sum.)2.NBT.B.5Fluently add and subtract within 100 using strategies based on place value (Adding the best hour sum and best minute sum, $9 + 14 = 23$.)
⭐ Fatten each spot on its own — the hour $9$ (not $12$!) plus the minutes $59$ give $9 + 5 + 9 = 23$.
⭐ Fatten each spot on its own — the hour $9$ (not $12$!) plus the minutes $59$ give $9 + 5 + 9 = 23$.
More like this
Same archetype — closest grade level first.