2018 AMC 10A Problem 22

From $\gcd(a,b)=24=2^3\cdot3$ and $\gcd(b,c)=36=2^2\cdot3^2$, the number $a$ is divisible by $2^3\cdot3$ but not by $3^2$. From $\gcd(b,c)=36$ and $\gcd(c,d)=54=2\cdot3^3$, the number $d$ is divisible by $2\cdot3^3$ but not by $2^2$. Therefore $\gcd(d,a)=2\cdot3\cdot n$, where $n$ has no factor $2$ or $3$. Since $70<6n<100,$ the integer $n$ satisfies $12\le n\le16.$ The only value in this range with no factor $2$ or $3$ is $13,$ so $13$ must divide $a.$ Thus, D is the correct answer.