2023 AMC 12B Problems/Problem 16

Problem

In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ 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 $x?$

$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\qquad\textbf{(E) }9$

Solution 1

This problem asks to find largest $x$ that cannot be written as $6 a + 10 b + 15 c = x, \hspace{1cm} (1)$

where $a, b, c \in \Bbb Z_+$ .

Denote by $r \in \left\{ 0, 1 \right\}$ the remainder of $x$ divided by 2. Modulo 2 on Equation (1), we get By using modulus $m \in \left\{ 2, 3, 5 \right\}$ on the equation above, we get $c \equiv r \pmod{2}$ .

Following from Chicken McNugget's Theorem, we have that any number that is no less than $(3-1)(5-1) = 8$ can be expressed in the form of $3a + 5b$ with $a, b \in \Bbb Z_+$ .

Therefore, all even numbers that are at least equal to $2 \cdot 8 + 15 \cdot 0 = 16$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ . All odd numbers that are at least equal to $2 \cdot 8 + 15 \cdot 1 = 31$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ .

The above two cases jointly imply that all numbers that are at least 30 can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ .

Next, we need to prove that 29 cannot be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$ .

Because 29 is odd, we must have $c \equiv 1 \pmod{2}$ . Because $a, b, c \in \Bbb Z_+$ , we must have $c = 1$ . Plugging this into Equation (1), we get $3 a + 5 b = 7$ . However, this equation does not have non-negative integer solutions.

All analysis above jointly imply that the largest $x$ that has no non-negative integer solution to Equation (1) is 29. Therefore, the answer is $2 + 9 = \boxed{\textbf{(D) 11}}$ .

~Steven Chen, www.professorchenedu.com

Solution 2

Arrange the positive integers into rows of $6$ , like this: $\begin{array}{|c|c|c|c|c|c|} \hline 01 & 02 & 03 & 04 & 05 & \textcolor{red}{\boxed{06}} \\ \hline 07 & 08 & 09 & \textcolor{red}{\boxed{10}} & 11 & \textcolor{red}{\boxed{12}} \\ \hline 13 & 14 & \textcolor{red}{\boxed{15}} & \textcolor{red}{\boxed{16}} & 17 & \textcolor{red}{\boxed{18}} \\ \hline 19 & \textcolor{red}{\boxed{20}} & \textcolor{red}{\boxed{21}} & \textcolor{red}{\boxed{22}} & 23 & \textcolor{red}{\boxed{24}} \\ \hline \textcolor{red}{\boxed{25}} & \textcolor{red}{\boxed{26}} & \textcolor{red}{\boxed{27}} & \textcolor{red}{\boxed{28}} & 29 & \textcolor{red}{\boxed{30}} \\ \hline \textcolor{red}{\boxed{31}} & \textcolor{red}{\boxed{32}} & \textcolor{red}{\boxed{33}} & \textcolor{red}{\boxed{34}} & \textcolor{red}{\boxed{35}} & \textcolor{red}{\boxed{36}} \\ \hline \end{array}$ $\vdots$ Observe that if any number can be made from a combination of $6$ s, $10$ s, and $15$ 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 $2+9= \boxed{\textbf{(D) 11}}$ .

(sorry for the bad formatting - feel free to edit)

~JN, ab_godder

Solution 3 (Modulo 6)

Let the number of $6$ cent coins be $a$ , the number of $10$ cent coins be $b$ , and the number of $15$ cent coins be $c$ . We get the Diophantine equation

$6a + 10b + 15c = k$

and we wish to find the largest possible value of $k$

Construct the following $\mod 6$ table of $6$ , $10$ , and $15$

$\begin{array}{c|ccc} & & & \\ \text{number of coins} & 6 & 10 & 15 \\ \hline & & & \\ 1 & 0 & 4 & 3\\ & & & \\ 2 & 0 & 2 & 0 \\ \end{array}$

There are only $6$ possible residues for $6$ , they are: $0$ , $1$ , $2$ , $3$ , $4$ , and $5$ .


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 $6$ , $10$ , and $15$ cent coins is $29$ , $2 + 9 = \boxed{\textbf{(D) 11}}$ .

~isabelchen

Solution 4

We claim that the largest number that cannot be obtained using $6$ , $10$ , and $15$ cent coins is $29$ .

Let's first focus on the combination of $6$ , $10$ . As both of them are even numbers, we cannot obtain any odd numbers from these two but requires $15$ to sum up to an odd number. Notice that by Chicken McNugget Theorem, the largest even number cannot be obtained by $6$ , $10$ is $2(3\cdot 5-3-5)=14$ . Add this with $29$ , we can easily verify that $29$ cannot be obtained by $6$ , $10$ , and $15$ as it needs at least one odd number, with the remaining part cannot be represented by $6$ and $10$ .

Let's show that any number greater than $29$ can be obtained. First, any even numbers greater than $29$ can be obtained by $6$ and $10$ by the Chicken McNugget Theorem. Next, any odd number greater than $29$ can be obtained by adding one $15$ with some $6$ s and $10$ s, which is also shown by the Chicken McNugget Theorem. This completes the proof. So the answer is $2+9 = \boxed{\textbf{(D) 11}}$

~ZZZIIIVVV

Solution 4 (apply Chicken McNugget Theorem twice)

First considering the two terms

$6a+10b = 2(3a+5b)$

$3a+5b = (3-1)(5-1)+t = 8+t, t \geq 0$

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 $2+9 = \boxed{\textbf{(D) 11}}$

~luckuso

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

https://youtu.be/E8WsGfn95mk

Video Solution 2

~Steven Chen, www.professorchenedu.com

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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