imc

The GCD information tells us that $24$ divides $a$, both $24$ and $36$ divide $b$, both $36$ and $54$ divide $c$, and $54$ divides $d$. Note that we have the prime factorizations: $$\begin{array}{rl}24& =2^3\cdot 3,\\ 36& =2^2\cdot 3^2,\\ 54& =2\cdot 3^3.\end{array}$$ Hence we have $$\begin{array}{rl}a& =2^3\cdot 3\cdot w\\ b& =2^3\cdot 3^2\cdot x\\ c& =2^2\cdot 3^3\cdot y\\ d& =2\cdot 3^3\cdot z\end{array}$$ for some positive integers $w,x,y,z$. Now if $3$ divides $w$, then $\gcd (a,b)$ would be at least $2^3\cdot 3^2$ which is too large, hence $3$ does not divide $w$. Similarly, if $2$ divides $z$, then $\gcd (c,d)$ would be at least $2^2\cdot 3^3$ which is too large, so $2$ does not divide $z$. Therefore, $$\gcd (a,d)=2\cdot 3\cdot \gcd (w,z)$$ where neither $2$ nor $3$ divide $\gcd (w,z)$. In other words, $\gcd (w,z)$ is divisible only by primes that are at least $5$. The only possible value of $\gcd (a,d)$ between $70$ and $100$ and which fits this criterion is $78=2\cdot 3\cdot 13$, so the answer is $\boxed{\text{ 13}}$.