View source for 2012 AMC 8 Problems/Problem 16

==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?

<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>

==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.

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

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

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

== 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 <math>97531</math> and <math>86420</math>. The digits can be interchangeable between numbers because we only care about the actual digits. 

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

The second digit must be either <math>7</math> or <math>6</math>. This doesn't cancel any choices.

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>. 

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

This leaves us with <math>\boxed{\textbf{(C)}\ 87431}</math>.

==See Also==
{{AMC8 box|year=2012|num-b=15|num-a=17}}
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