2012 AMC 8 Problems/Problem 16
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 | |
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