AMC8 2026

Let $x$ be the number of gold coins, and let $y$ be the number of silver coins. We know that $y$ must be $0$, $1$, and $2$, since if $y$ was $3$ or more, then the stack of coins would be more than $8$ mm tall. Now we can use casework to find all possible stacks. Case $1$: If $y=0$, then there is only one way to make the stack of coins, which is $8$ gold coins. Case $2$: If $y=1$, then there must be $5$ gold coins, since $8-3\cdot1=5$. There are $6$ coins in total, and we can put the silver coin anywhere in the stack, so there are $\binom{6}{1}=6$ ways to place the silver coin, giving $6$ different possible stacks with $1$ silver coin. (Note that when you place the silver coin anywhere, the rest of the stack is predetermined, because the rest of the stack must be gold, so we only have to account for one type of coin, in this case, the silver coin.) Case $3$: If $y=2$, then there must be $2$ gold coins, so we have 4 coins in total. We can distribute the $2$ silver coins in $\binom{4}{2}=6$ different ways, giving $6$ possible stacks with $2$ silver coins. (Again, note that wherever you place the silver coins, the rest of the stack is predetermined, because the rest of the stack must be gold, so we only have to account for one type of coin, in this case, the silver coin.) Adding all the possible number of cases in all $3$ cases gives $1+6+6=13$, thus our answer is $\boxed{\textbf{(D)}\ 13}$