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Difference between revisions of "2012 AMC 8 Problems/Problem 16"
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==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?
 
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>
 
<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 1==
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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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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>.
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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>.
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From the estimations, we can say that the answer to this problem is <math> \boxed{\textbf{(C)}\ 87431} </math>.
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p.s. USE INTUITION, see answer choices before solving any question -litttle_master
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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>3</math> or <math>2</math>. This cancels out <math>\textbf{(E)}\ 97403</math>.
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This leaves us with <math>\boxed{\textbf{(C)}\ 87431}</math>.
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== Solution 3 ==
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If we use intuition, we know that the digits will be IN ORDER, to maximze the number. This eliminates <math>\textbf{(B)}</math>, <math>\textbf{(D)}</math>,and <math>\textbf{(E)}</math>. Additionally, the 2 two numbers must have 9 and 8 for the first digit to maximize the sum, eliminating <math>\textbf{(A)}</math>. This leaves <math>\boxed{\textbf{(C)}\ 87431}</math>.
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-written by litttle_master purely, not copied from anywhere
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== Video Solution by OmegaLearn==
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https://youtu.be/HISL2-N5NVg?t=654
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~ pi_is_3.14
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https://youtu.be/trAjltkbSWo ~savannahsolver
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https://www.youtube.com/watch?v=dQw4w9WgXcQ
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=15|num-a=17}}
 
{{AMC8 box|year=2012|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 14:49, 27 December 2023

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 $\boxed{\textbf{(C)}\ 87431}$.

p.s. USE INTUITION, see answer choices before solving any question -litttle_master

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 $\boxed{\textbf{(C)}\ 87431}$.

Solution 3

If we use intuition, we know that the digits will be IN ORDER, to maximze the number. This eliminates $\textbf{(B)}$, $\textbf{(D)}$,and $\textbf{(E)}$. Additionally, the 2 two numbers must have 9 and 8 for the first digit to maximize the sum, eliminating $\textbf{(A)}$. This leaves $\boxed{\textbf{(C)}\ 87431}$. -written by litttle_master purely, not copied from anywhere

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=654

~ pi_is_3.14

https://youtu.be/trAjltkbSWo ~savannahsolver

https://www.youtube.com/watch?v=dQw4w9WgXcQ

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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