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

m
 
(32 intermediate revisions by 11 users not shown)
Line 1: Line 1:
#REDIRECT [[2006 AMC 12A Problems\Problem 4]]
+
{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2006 AMC 10A #4]]}}
 +
== Problem ==
 +
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
 +
 
 +
<math>\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\  23</math>
 +
 
 +
== Solution 1 ==
 +
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>
 +
 
 +
==Solution 2 ([[matrix]]) ==
 +
 
 +
With a matrix, we can see
 +
<math>
 +
\begin{bmatrix}
 +
1+2&9&6&3\\
 +
1+1&8&5&2\\
 +
1+0&7&4&1
 +
\end{bmatrix}
 +
</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>
 +
 
 +
==Solution 3==
 +
 
 +
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 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>.
 +
 
 +
~[[User:Dairyqueenxd|Dairyqueenxd]]
 +
 
 +
== See also ==
 +
{{AMC12 box|year=2006|ab=A|num-b=3|num-a=5}}
 +
{{AMC10 box|year=2006|ab=A|num-b=3|num-a=5}}
 +
{{MAA Notice}}
 +
 
 +
[[Category:Introductory Number Theory Problems]]

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 (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

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