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

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== Solution ==
 
== Solution ==
  
From the [[greedy algorithm]], we have 9 in the hours section and 59 in the minutes section. <math>9+5+9=19 \Rightarrow \mathrm {(E)}</math>
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From the [[greedy algorithm]], we have 9 in the hours section and 59 in the minutes section. <math>9+5+9=23 \Rightarrow \mathrm {(E)}</math>
  
 
== See also ==
 
== See also ==

Revision as of 14:46, 3 November 2007

Problem

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?

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

$\mathrm{(E) \ }  23$

Solution

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

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
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