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

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Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2006 AMC 10A #4]]}}
<center><math> \mathrm{(A) \ } -h\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } h\qquad \mathrm{(D) \ } 2h\qquad \mathrm{(E) \ } h^3 </math></center>
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== Problem ==
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A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
  
== Solution==
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<math>\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\  23</math>
Plugging our values in the function, we have
 
  
<center><math>\displaystyle h^3-h</math></center>
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== Solution 1 ==
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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>
  
Plugging in the function once more, we have
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==Solution 2 ([[matrix]]) ==
  
<center><math>\displaystyle h^3-(h^3-h)=h,  
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With a matrix, we can see
(C)</math></center>
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<math>
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\begin{bmatrix}
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1+2&9&6&3\\
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1+1&8&5&2\\
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1+0&7&4&1
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\end{bmatrix}
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</math>
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The largest single digit sum we can get is <math>9</math>.
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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>
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==Solution 3==
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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>.
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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>.
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Adding these up, we get <math>9+5+9=\boxed{\textbf{(E) }23}</math>.
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~[[User:Dairyqueenxd|Dairyqueenxd]]
  
 
== See also ==
 
== See also ==
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{{AMC12 box|year=2006|ab=A|num-b=3|num-a=5}}
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{{AMC10 box|year=2006|ab=A|num-b=3|num-a=5}}
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{{MAA Notice}}
  
* [[2006 AMC 12A Problems]]
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[[Category:Introductory Number Theory Problems]]

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?

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

Solution 1

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

Solution 2 (matrix)

With a matrix, we can see $\begin{bmatrix} 1+2&9&6&3\\ 1+1&8&5&2\\ 1+0&7&4&1 \end{bmatrix}$ The largest single digit sum we can get is $9$. For the minutes digits, we can combine the largest $2$ digits, which are $9,5 \Rightarrow 9+5=14$, and finally $14+9=\boxed{\textbf{(E) }23}$

Solution 3

We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$. Therefore, the maximum value of the digits for the hours is when the watch displays $09$, which gives us $0+9=9$.

Next, we look at the value of the minutes section, which varies from $00-59$. Let this value be a number $ab$. We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$.

Adding these up, we get $9+5+9=\boxed{\textbf{(E) }23}$.

~Dairyqueenxd

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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