Difference between revisions of "2023 AMC 12B Problems/Problem 16"
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+ | ==Problem== | ||
+ | In the state of Coinland, coins have values <math>6,10,</math> and <math>15</math> cents. Suppose <math>x</math> 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 <math>x?</math> | ||
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+ | <math>\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\qquad\textbf{(E) }9</math> | ||
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==Solution== | ==Solution== | ||
Revision as of 18:39, 15 November 2023
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
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 MuNugget'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)