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Difference between revisions of "2006 AMC 12A Problems/Problem 4"
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2008 AMC 10A #4]]}}
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2006 AMC 10A #4]]}}
 
== Problem ==
 
== Problem ==
 
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
 
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
  
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<math>\mathrm{(A)}\ 17\qquad\mathrm{(B)}\ 19\qquad\mathrm{(C)}\ 21\qquad\mathrm{(D)}\ 22\qquad\mathrm{(E)}\  23</math>
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<math>\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\  23</math>
  
 
== Solution 1 ==
 
== Solution 1 ==
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From the [[greedy algorithm]], we have <math>9</math> in the hours section and <math>59</math> in the minutes section. <math>9+5+9=23\Rightarrow\mathrm{(E)}</math>
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From the [[greedy algorithm]], we have <math>9</math> in the hours section and <math>59</math> in the minutes section. <math>9+5+9=\boxed{\textbf{(E) }23}</math>
  
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==Solution 2==
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==Solution 2 ([[matrix]]) ==
  
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With a matrix we can see
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With a matrix, we can see
 
<math>
 
<math>
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\[\begin{bmatrix}
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\begin{bmatrix}
 
1+2&9&6&3\\
 
1+2&9&6&3\\
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1+11&8&5&2\\
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1+1&8&5&2\\
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1+0&7&4&2\\
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1+0&7&4&1
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\end{bmatrix}\]
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\end{bmatrix}
 
</math>
 
</math>
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The largest digit sum we can see is <math>9</math>
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The largest single digit sum we can get is <math>9</math>.
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For the minutes digits, we can combine the largest <math>2</math> digits, which are <math>9,5=9+5=14</math> which we can then do <math>14+9=23</math>
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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 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>.
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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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 +
~[[User:Dairyqueenxd|Dairyqueenxd]]
  
 
== See also ==
 
== See also ==

Latest revision as of 10:20, 2 January 2022

The following problem is from both the 2006 AMC 12A #4 and 2006 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 $9+5+9=\boxed{\textbf{(E) }23}$.

~Dairyqueenxd

See also

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

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