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

Revision as of 19:00, 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 watch displays time in AM and PM, the value for the hours section varies from $00-12$ . Therefore, the maximum value of the digits for the hours is when the watch displays $09$ , which gives us $0+9=9$ .

Next, we look at the value of the minutes section, which varies from $00-59$ . Let this value be a number $ab$ . We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$ .

Adding these up, we get $9+5+9=\boxed{\textbf{(E) }23}$ .

2006 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 23: / Line 23: @@
 ==Solution 3==
-We first note that since the watch displays time in AM and PM, the value for the hours varies from <math>00-12</math>. Therefore, the maximum value of the digits for the hours is when the watch displays <math>09</math>, which gives us <math>0+9=9</math>.
+We first note that since the watch displays time in AM and PM, the value for the hours section varies from <math>00-12</math>. Therefore, the maximum value of the digits for the hours is when the watch displays <math>09</math>, which gives us <math>0+9=9</math>.
-Next, we look at the value of minutes, which varies from <math>00-59</math>. Let this value be a number <math>ab</math>. We quickly find that the maximum value for <math>a</math> and <math>b</math> is respectively <math>5</math> and <math>9</math>.
+Next, we look at the value of the minutes section, which varies from <math>00-59</math>. Let this value be a number <math>ab</math>. We quickly find that the maximum value for <math>a</math> and <math>b</math> is respectively <math>5</math> and <math>9</math>.
 Adding these up, we get <math>9+5+9=\boxed{\textbf{(E) }23}</math>.

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