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

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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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p.s. USE INTUITION, see answer choices before solving any question -litttle_master
  
 
== Solution 2 ==
 
== Solution 2 ==
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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>.  
 
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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The fourth digit must be <math>3</math> or <math>2</math>. This cancels out <math>\textbf{(E)}\ 97403</math>.  
  
 
This leaves us with <math>\boxed{\textbf{(C)}\ 87431}</math>.
 
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}}
 
{{MAA Notice}}
 
{{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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