2023 AMC 12B Problems/Problem 16
Contents
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
Column 3 - cross out 15
Column 4 - cross out 10
Column 5 - cross out 35
Column 6 - cross all out.
The maximum number that remains is 29. Answer is 2+9=.
(sorry for the bad formatting - feel free to edit)
~JN
~format edit by ab_godder
Solution 3 (Modulo 6)
Let the number of cent coins be , the number of cent coins be , and the number of cent coins be . We get the Diophantine equation
and we wish to find the largest possible value of
Construct the following table of , , and
There are only possible residues for , they are: , , , , , and .
We can obtain any value that is because we have cent coins.
We can obtain values that equal by using one cent coin, one cent coin, and as many cent coins needed. The smallest value that equals we can obtain is . Therefore, the largest value that equals we cannot obtain is
We can obtain values that equal by using two cent coins, and as many cent coins as needed. The smallest value that equals we can obtain is . Therefore, the largest value that equals we cannot obtain is
We can obtain values that equal by using one cent coin, and as many cent coins as needed. The smallest value that equals we can obtain is . Therefore, the largest value that equals we cannot obtain is
We can obtain values that equal by using one cent coin, and as many cent coins as needed. The smallest value that equals we can obtain is . Therefore, the largest value that equals we cannot obtain is
We can obtain values that equal by using two cent coins, one cent coin, and as many cent coins as needed. The smallest value that equals we can obtain is . Therefore, the largest value that equals we cannot obtain is
Hence, the largest value in cents we cannot obtain using , , and cent coins is , .
Solution 4
We claim that the largest number that cannot be obtained using , , and cent coins is .
Let's first focus on the combination of , . As both of them are even numbers, we cannot obtain any odd numbers from these two but requires to sum up to an odd number. Notice that by Chicken McNugget Theorem, the largest even number cannot be obtained by , is . Add this with , we can easily verify that cannot be obtained by , , and as it needs at least one odd number, with the remaining part cannot be represented by and .
Let's show that any number greater than can be obtained. First, any even numbers greater than can be obtained by and by the Chicken McNugget Theorem. Next, any odd number greater than can be obtained by adding one with some s and s, which is also shown by the Chicken McNugget Theorem. This completes the proof. So the answer is
~ZZZIIIVVV
Solution 4 (apply Chicken McNugget Theorem twice)
First considering the two terms
6a+10b = 2(3a+5b)
3a+5b = (3-1)(5-1)+t = 8+t, t>=0
Again applying the two variable formula for the terms in brackets we see that
6a+10b+15c = 2(8+t) + 15c = 2t +15c + 16 = (2-1)(15-1) + s+ 16 = 30+s, s>=0
so the given expression 6a+10b+15c produces all numbers from 30 and 29 is the largest number that could not be produced,so the answer is
- Note: Chicken McNugget Theorem: any positive integer N= (a-1)(b-1)+t ( a,b co-prime, t>=0) can always be represented as N = aP+ bQ ( p,q are non-negative integer)
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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