Difference between revisions of "2012 AMC 8 Problems/Problem 16"
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This leaves us with <math>\boxed{\textbf{(C)}\ 87431}</math>. | This leaves us with <math>\boxed{\textbf{(C)}\ 87431}</math>. | ||
− | == Video Solution == | + | == Video Solution by OmegaLearn== |
https://youtu.be/HISL2-N5NVg?t=654 | https://youtu.be/HISL2-N5NVg?t=654 | ||
Revision as of 18:04, 31 December 2022
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 .
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 .
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=654
~ pi_is_3.14
https://youtu.be/trAjltkbSWo ~savannahsolver
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.