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> | ||
− | ==Solution== | + | ==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. | 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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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>. | ||
+ | |||
+ | 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 <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>3</math> or <math>2</math>. This cancels out <math>\textbf{(E)}\ 97403</math>. | ||
+ | |||
+ | This leaves us with <math>\boxed{\textbf{(C)}\ 87431}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | 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== | ||
+ | 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== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=15|num-a=17}} | {{AMC8 box|year=2012|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:49, 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?
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: and
. To determine the answer we will have to use estimation and the first two digits of the numbers.
For the number that would maximize the sum would start with
. The first two digits of
(when rounded) are
. Adding
and
, we find that the first three digits of the sum of the two numbers would be
.
For the number that would maximize the sum would start with
. The first two digits of
(when rounded) are
. Adding
and
, we find that the first three digits of the sum of the two numbers would be
.
From the estimations, we can say that the answer to this problem is .
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 and
. The digits can be interchangeable between numbers because we only care about the actual digits.
The first digit must be either or
. This immediately knocks out
.
The second digit must be either or
. This doesn't cancel any choices.
The third digit must be either or
. This knocks out
and
.
The fourth digit must be or
. This cancels out
.
This leaves us with .
Solution 3
If we use intuition, we know that the digits will be IN ORDER, to maximze the number. This eliminates ,
,and
. Additionally, the 2 two numbers must have 9 and 8 for the first digit to maximize the sum, eliminating
. This leaves
.
-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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.