AMC12 2023 B

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 the Chicken McNugget 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 $2 + 9 = \boxed{\textbf{(D) 11}}$.