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| | + | {{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? | ||
− | <math>\ | + | <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 == | ||
− | 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= | + | 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== | + | ==Solution 2 ([[matrix]]) == |
− | With a matrix we can see | + | With a matrix, we can see |
<math> | <math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
1+2&9&6&3\\ | 1+2&9&6&3\\ | ||
− | 1+ | + | 1+1&8&5&2\\ |
− | 1+0&7&4& | + | 1+0&7&4&1 |
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
− | The largest digit sum we can | + | 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> which we | + | 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 == | == See also == |
Latest revision as of 11:20, 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?
Solution 1
From the greedy algorithm, we have in the hours section and
in the minutes section.
Solution 2 (matrix)
With a matrix, we can see
The largest single digit sum we can get is
.
For the minutes digits, we can combine the largest
digits, which are
, and finally
Solution 3
We first note that since the watch displays time in AM and PM, the value for the hours section varies from . Therefore, the maximum value of the digits for the hours is when the watch displays
, which gives us
.
Next, we look at the value of the minutes section, which varies from . Let this value be a number
. We quickly find that the maximum value for
and
is respectively
and
.
Adding these up, we get .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.