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2023 AMC 12B Problems/Problem 16

Revision as of 22:07, 19 August 2024 by Cyantist (talk | contribs) (→‎Solution 4)

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 (Professor Chen Education Palace, 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

~format edit by 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 $5$. 

$0 \mod 6$
We can obtain any value that is $0 \mod 6$ because we have $6$ cent coins.

$1 \mod 6$
We can obtain values that equal $1 \mod 6$ by using one $10$ cent coin, one $15$ cent coin, and as many $6$ cent coins needed.
The smallest value that equals $1 \mod 6$ we can obtain is $10+15 = 25$. Therefore, the largest value that equals $1 \mod 6$ we cannot obtain is $25-6=19$

$2 \mod 6$
We can obtain values that equal $2 \mod 6$ by using two $10$ cent coins, and as many $6$ cent coins as needed.
The smallest value that equals $2 \mod 6$ we can obtain is $10+10 = 20$. Therefore, the largest value that equals $2 \mod 6$ we cannot obtain is $20-6=14$

$3 \mod 6$
We can obtain values that equal $3 \mod 6$ by using one $15$ cent coin, and as many $6$ cent coins as needed.
The smallest value that equals $3 \mod 6$ we can obtain is $15$. Therefore, the largest value that equals $3 \mod 6$ we cannot obtain is $15-6=9$

$4 \mod 6$
We can obtain values that equal $4 \mod 6$ by using one $10$ cent coin, and as many $6$ cent coins as needed.
The smallest value that equals $4 \mod 6$ we can obtain is $10$. Therefore, the largest value that equals $4 \mod 6$ we cannot obtain is $10-6=4$

$5 \mod 6$
We can obtain values that equal $5 \mod 6$ by using two $10$ cent coins, one $15$ cent coin, and as many $6$ cent coins as needed.
The smallest value that equals $5 \mod 6$ we can obtain is $10+10+15 = 35$. Therefore, the largest value that equals $5 \mod 6$ we cannot obtain is $35-6=29$

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>=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) + 16 = 30+s, s>0

so 29 is the largest number that could not represented by 6a+10b+15c

~luckuso

Video Solution 1 by OmegaLearn

https://youtu.be/E8WsGfn95mk

Video Solution

https://youtu.be/T5nqrNGvf4Q

~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
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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