2023 AMC 12B Problems/Problem 16
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[hide]Problem
In the state of Coinland, coins have values and cents. Suppose is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of
Solution 1
This problem asks to find largest that cannot be written as where .
Denote by the remainder of divided by 2. Modulo 2 on Equation (1), we get By using modulus on the equation above, we get .
Following from Chicken McNugget's theorem, we have that any number that is no less than can be expressed in the form of with .
Therefore, all even numbers that are at least equal to can be written in the form of Equation (1) with . All odd numbers that are at least equal to can be written in the form of Equation (1) with .
The above two cases jointly imply that all numbers that are at least 30 can be written in the form of Equation (1) with .
Next, we need to prove that 29 cannot be written in the form of Equation (1) with .
Because 29 is odd, we must have . Because , we must have . Plugging this into Equation (1), we get . However, this equation does not have non-negative integer solutions.
All analysis above jointly imply that the largest that has no non-negative integer solution to Equation (1) is 29. Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
Arrange the positive integers into rows of 6, like this:
Observe that if any number can be made from a combination of s, s, and s, then every number below it in the same column must also be possible to make, by simply adding 6s.
Thus, we will cross out any numbers that CAN be made as well as all numbers below it.
In column 1, 25 is possible (10+15) and so is everything below 25.
Column 2 - cross out 20 (10+10) and everything below it
Column 3 - cross out 15 and everything below it
Column 4 - cross out 10 and everything below it
Column 5 - cross out 35 (15+10+10) and everything below it
Column 6 - all numbers here are possible, so cross all out.
The maximum number that remains is 29. Answer is 2+9=11.
(sorry for the bad formatting - feel free to edit)
~JN
Solution 3 (mod 6)
There are only possible residues for , they are: , , , , , and .
We can obtain any integer that is because we have cent coins.
We can obtain integers that equal by using one cent coin, one cent coin, and as many cent coins that are needed. The smallest integer that equals we can obtain is . Therefore, the largest integer that equals we cannot obtain is
We can obtain integers that equal by using two cent coins, and as many cent coins that are needed. The smallest integer that equals we can obtain is . Therefore, the largest integer that equals we cannot obtain is
We can obtain integers that equal by using one cent coin, and as many cent coins that are needed. The smallest integer that equals we can obtain is . Therefore, the largest integer that equals we cannot obtain is
We can obtain integers that equal by using one cent coin, and as many cent coins that are needed. The smallest integer that equals we can obtain is . Therefore, the largest integer that equals we cannot obtain is
We can obtain integers that equal by using two cent coins, one cent coin, and as many cent coins that are needed. The smallest integer that equals we can obtain is . Therefore, the largest integer that equals we cannot obtain is
Hence, the largest integer we cannot obtain is , .
Video Solution 1 by OmegaLearn
See Also
2023 AMC 12B (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 AMC 12 Problems and Solutions |
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