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

m (Solution 2)
m (Solution 2)
Line 12: Line 12:
 
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+11&8&5&2\
 
1+11&8&5&2\
 
1+0&7&4&2
 
1+0&7&4&2
\end{bmatrix}\]
+
\end{bmatrix}
 
</math>
 
</math>
 
The largest digit sum we can see is <math>9</math>
 
The largest digit sum we can see is <math>9</math>

Revision as of 17:58, 23 July 2018

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?

$\mathrm{(A)}\ 17\qquad\mathrm{(B)}\ 19\qquad\mathrm{(C)}\ 21\qquad\mathrm{(D)}\ 22\qquad\mathrm{(E)}\  23$

Solution 1

From the greedy algorithm, we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=23\Rightarrow\mathrm{(E)}$

Solution 2

With a matrix we can see $\begin{bmatrix} 1+2&9&6&3\\ 1+11&8&5&2\\ 1+0&7&4&2 \end{bmatrix}$ The largest digit sum we can see is $9$ For the minutes digits, we can combine the largest $2$ digits, which are $9,5=9+5=14$ which we can then do $14+9=23$

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

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