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?

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

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 $3$ or $2$. This cancels out $\textbf{(E)}\ 97403$.

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

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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