Difference between revisions of "2006 AMC 12A Problems/Problem 4"

(Solution 2 (matrix))
(Solution 2 (matrix))
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The largest single digit sum we can get is <math>9</math>.
 
The largest single digit sum we can get is <math>9</math>.
 
For the minutes digits, we can combine the largest <math>2</math> digits, which are <math>9,5 \Rightarrow 9+5=14</math>, and finally <math>14+9=\boxed{\textbf{(E) }23}</math>
 
For the minutes digits, we can combine the largest <math>2</math> digits, which are <math>9,5 \Rightarrow 9+5=14</math>, and finally <math>14+9=\boxed{\textbf{(E) }23}</math>
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==Solution 3==
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We first note that since the hours are in AM and PM.
  
 
== See also ==
 
== See also ==

Revision as of 17:46, 16 December 2021

The following problem is from both the 2006 AMC 12A #4 and 2008 AMC 10A #4, so both problems redirect to this page.

Problem

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?

$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\  23$

Solution 1

From the greedy algorithm, we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\boxed{\textbf{(E) }23}$

Solution 2 (matrix)

With a matrix, we can see $\begin{bmatrix} 1+2&9&6&3\\ 1+1&8&5&2\\ 1+0&7&4&1 \end{bmatrix}$ The largest single digit sum we can get is $9$. For the minutes digits, we can combine the largest $2$ digits, which are $9,5 \Rightarrow 9+5=14$, and finally $14+9=\boxed{\textbf{(E) }23}$

Solution 3

We first note that since the hours are in AM and PM.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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