Difference between revisions of "2006 AMC 12A Problems/Problem 4"
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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> | <math>\mathrm{(A)}\ 17\qquad\mathrm{(B)}\ 19\qquad\mathrm{(C)}\ 21\qquad\mathrm{(D)}\ 22\qquad\mathrm{(E)}\ 23</math> | ||
| − | == Solution == | + | == 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=23\Rightarrow\mathrm{(E)}</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=23\Rightarrow\mathrm{(E)}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | With a matrix we can see | ||
| + | <math> | ||
| + | \[\begin{bmatrix} | ||
| + | 1+2&9&6&3\\ | ||
| + | 1+11&8&5&2\\ | ||
| + | 1+0&7&4&2\\ | ||
| + | \end{bmatrix}\] | ||
| + | </math> | ||
| + | The largest digit sum we can see is <math>9</math> | ||
| + | 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> | ||
== See also == | == See also == | ||
Revision as of 20:55, 27 January 2018
- The following problem is from both the 2006 AMC 12A #4 and 2008 AMC 10A #4, so both problems redirect to this page.
Contents
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
With a matrix we can see
$\[\begin{bmatrix}
1+2&9&6&3\\
1+11&8&5&2\\
1+0&7&4&2\\
\end{bmatrix}\]$ (Error compiling LaTeX. Unknown error_msg)
The largest digit sum we can see is
For the minutes digits, we can combine the largest 
digits, which are 
which we can then do 
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. 
