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

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

@@ Line 12: / Line 12: @@
 From the estimations, we can say that the answer to this problem is <math> \boxed{\textbf{(C)}\ 87431} </math>.
+p.s. USE INTUITION, see answer choices before solving any question -litttle_master
 == Solution 2 ==
@@ Line 30: / Line 32: @@
 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>.
+-written by litttle_master purely, not copied from anywhere
 == Video Solution by OmegaLearn==

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

Page

Toolbox

Search