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Difference between revisions of "2023 AMC 12B Problems/Problem 16"
(→‎Solution 3 (mod 6))
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<cmath>6a+10b+15c</cmath>
 
<cmath>6a+10b+15c</cmath>
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−
 
−
 
 
<cmath>\begin{array}{c|ccc}
 
<cmath>\begin{array}{c|ccc}
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  &  &  & \\ [-2ex]
+
  &  &  & \\
−
n & 6 & 10 & 15 \\ [1ex]
+
n & 6 & 10 & 15 \\  
 
\hline
 
\hline
−
  &  &  & \\ [-1ex]
+
  &  &  & \\  
 
1 & 0 & 4 & 3\\
 
1 & 0 & 4 & 3\\
 
  &  &  &  \\
 
  &  &  &  \\
 
2 & 0 & 2 & 0 \\
 
2 & 0 & 2 & 0 \\
−
&  &  & \\
 
−
3 & 0 & 0 & 3 \\ [1ex]
 
 
\end{array}</cmath>
 
\end{array}</cmath>
  
−
(Writing in progress......)
+
There are only <math>6</math> possible residues for <math>6</math>, they are: <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math>.
 +
 
 +
<math>0 \mod 6</math>
 +
We can obtain any integer that is <math>0 \mod 6</math> because we have <math>6</math> cent coins.
 +
 
 +
<math>1 \mod 6</math>
 +
We can obtain integers that equal <math>1 \mod 6</math> by using one <math>10</math> cent coin, one <math>15</math> cent coin, and as many <math>6</math> cent coins that are needed.
 +
The smallest integer that equals <math>1 \mod 6</math> we can obtain is <math>10+15 = 25</math>. Therefore, the largest integer that equals <math>1 \mod 6</math> we cannot obtain is <math>25-6=19</math>
 +
 
 +
<math>2 \mod 6</math>
 +
We can obtain integers that equal <math>2 \mod 6</math> by using two <math>10</math> cent coins, and as many <math>6</math> cent coins that are needed.
 +
The smallest integer that equals <math>2 \mod 6</math> we can obtain is <math>10+10 = 20</math>. Therefore, the largest integer that equals <math>2 \mod 6</math> we cannot obtain is <math>20-6=14</math>
 +
 
 +
<math>3 \mod 6</math>
 +
We can obtain integers that equal <math>3 \mod 6</math> by using one <math>15</math> cent coin, and as many <math>6</math> cent coins that are needed.
 +
The smallest integer that equals <math>3 \mod 6</math> we can obtain is <math>15</math>. Therefore, the largest integer that equals <math>3 \mod 6</math> we cannot obtain is <math>15-6=9</math>
 +
 
 +
<math>4 \mod 6</math>
 +
We can obtain integers that equal <math>4 \mod 6</math> by using one <math>10</math> cent coin, and as many <math>6</math> cent coins that are needed.
 +
The smallest integer that equals <math>4 \mod 6</math> we can obtain is <math>10</math>. Therefore, the largest integer that equals <math>4 \mod 6</math> we cannot obtain is <math>10-6=4</math>
 +
 
 +
<math>5 \mod 6</math>
 +
We can obtain integers that equal <math>5 \mod 6</math> by using two <math>10</math> cent coins, one <math>15</math> cent coin, and as many <math>6</math> cent coins that are needed.
 +
The smallest integer that equals <math>5 \mod 6</math> we can obtain is <math>10+10+15 = 35</math>. Therefore, the largest integer that equals <math>5 \mod 6</math> we cannot obtain is <math>35-6=29</math>
 +
 
 +
Hence, the largest integer we cannot obtain is <math>29</math>, <math>2 + 9 = \boxed{\textbf{(D) 11}}</math>.
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]

Revision as of 22:56, 15 November 2023

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{align*} 01, 02, 03, 04, 05, 06 \\ 07, 08, 09, 10, 11, 12 \\ 13, 14, 15, 16, 17, 18 \\ 19, 20, 21, 22, 23, 24 \\ 25, 26, 27, 28, 29, 30 \\ 31, 32, 33, 34, 35, 36 \\ ... \end{align*}

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 (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)

Let the number of $6$ cent coins be $a$, $10$ cent coins be $b$, $15$ cent coins be $c$.

\[6a+10b+15c\] \[\begin{array}{c|ccc} & & & \\ n & 6 & 10 & 15 \\ \hline & & & \\ 1 & 0 & 4 & 3\\ & & & \\ 2 & 0 & 2 & 0 \\ \end{array}\]

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

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

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

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

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

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

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

Hence, the largest integer we cannot obtain is $29$, $2 + 9 = \boxed{\textbf{(D) 11}}$.

~isabelchen

Solution 4 (Fast)

Its easy to see the smallest value $1 \pmod 6$ we can achieve is $25$, the smallest value $2 \pmod 6$ is $20$, the smallest value $3 \pmod 6$ is $15$, the smallest value $4 \pmod 6$ is $10$, and the smallest value $5\pmod 6$ is $35$. After each of these values, we are able to reach all larger values with the same residue. This implies that the last value we can't reach is $35-6 = 29 \implies \boxed{\textbf{(D) }}$

~AtharvNaphade

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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