2006 AMC 10A Problem 4

The hour part and the minute part are independent, so maximize each on its own.

Hours run from $1$ to $12$. A one-digit hour contributes its own value, at most $9$; the two-digit hours $10$, $11$, $12$ contribute only $1$, $2$, $3$. So the best hour is $9$, contributing $9$.

Minutes run from $00$ to $59$. The tens digit is at most $5$ and the ones digit at most $9$, and $59$ achieves both, contributing $5 + 9 = 14$.

The display $9{:}59$ therefore has digit sum $9 + 14 = 23$, and nothing larger is possible.

The answer is $\boxed{\textbf{(E)}\ 23}$.

When a quantity splits into independent pieces, the maximum of the sum is the sum of the maxima, provided each maximum is attainable at the same time (here $9{:}59$ is a real display). The twist is that "biggest number" and "biggest digit sum" differ: $12$ is a larger hour than $9$ but has a far smaller digit sum.

• Assuming the largest time $12{:}59$ gives the largest digit sum; $1 + 2 + 5 + 9 = 17$ (choice (A)) loses to $9{:}59$.
• Letting the minute tens digit be $9$ (as in $9{:}99$), which is not a valid time.