Difference between revisions of "2006 AMC 12A Problems/Problem 4"
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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> | 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 | + | With a matrix we can see |
<math> | <math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| Line 19: | Line 19: | ||
</math> | </math> | ||
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{( | + | 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>. |
== See also == | == See also == | ||
Revision as of 17:44, 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?
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 
.
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. 
