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Difference between revisions of "2012 AMC 8 Problems/Problem 16"

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<math> \textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)}\hspace{.05in}96240\qquad\textbf{(E)}\hspace{.05in}97403 </math>
 
<math> \textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)}\hspace{.05in}96240\qquad\textbf{(E)}\hspace{.05in}97403 </math>
  
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==Solution==
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==Solution 1==
 
In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choices with this property: <math> 76531 </math> and <math> 87431 </math>. To determine the answer we will have to use estimation and the first two digits of the numbers.
 
In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choices with this property: <math> 76531 </math> and <math> 87431 </math>. To determine the answer we will have to use estimation and the first two digits of the numbers.
  
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From the estimations, we can say that the answer to this problem is <math> \boxed{\textbf{(C)}\ 87431} </math>.
 
From the estimations, we can say that the answer to this problem is <math> \boxed{\textbf{(C)}\ 87431} </math>.
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== Solution 2 ==
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In order to determine the largest number possible, we have to evenly distribute the digits when adding. The two numbers that show an example of this are <math>97531</math> and <math>86420</math>. The digits can be interchangeable between numbers because we only care about the actual digits.
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The first digit must be either <math>9</math> or <math>8</math>. This immediately knocks out <math>\textbf{(A)}\ 76531</math>.
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The second digit must be either <math>7</math> or <math>6</math>. This doesn't cancel any choices.
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The third digit must be either <math>5</math> or <math>4</math>. This knocks out <math>\textbf{(B)}\ 86724</math> and <math>\textbf{(D)}\ 96240</math>.
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The fourth digit must be <math>4</math> or <math>3</math>. This cancels out <math>\textbf{(E)}\ 97403</math>.
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This leaves us with <math>\boxed{\textbf{(C)}\ 87431}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=15|num-a=17}}
 
{{AMC8 box|year=2012|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:28, 9 November 2013

Problem

Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?

$\textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)}\hspace{.05in}96240\qquad\textbf{(E)}\hspace{.05in}97403$

Solution 1

In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choices with this property: $76531$ and $87431$. To determine the answer we will have to use estimation and the first two digits of the numbers.

For $76531$ the number that would maximize the sum would start with $98$. The first two digits of $76531$ (when rounded) are $77$. Adding $98$ and $77$, we find that the first three digits of the sum of the two numbers would be $175$.

For $87431$ the number that would maximize the sum would start with $96$. The first two digits of $87431$ (when rounded) are $87$. Adding $96$ and $87$, we find that the first three digits of the sum of the two numbers would be $183$.

From the estimations, we can say that the answer to this problem is $\boxed{\textbf{(C)}\ 87431}$.

Solution 2

In order to determine the largest number possible, we have to evenly distribute the digits when adding. The two numbers that show an example of this are $97531$ and $86420$. The digits can be interchangeable between numbers because we only care about the actual digits.

The first digit must be either $9$ or $8$. This immediately knocks out $\textbf{(A)}\ 76531$.

The second digit must be either $7$ or $6$. This doesn't cancel any choices.

The third digit must be either $5$ or $4$. This knocks out $\textbf{(B)}\ 86724$ and $\textbf{(D)}\ 96240$.

The fourth digit must be $4$ or $3$. This cancels out $\textbf{(E)}\ 97403$.

This leaves us with $\boxed{\textbf{(C)}\ 87431}$.

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 AJHSME/AMC 8 Problems and Solutions

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