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? | ||
| Line 23: | Line 23: | ||
==Solution 3== | ==Solution 3== | ||
| − | We first note that since the | + | 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 10:20, 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. 
