smc

Denote by ${\nu }_p(x)$ the number of prime factor $p$ in number $x$. We index Equations given in this problem from (1) to (7). First, we compute ${\nu }_2(x)$ for $x\in \left\{a,b,c,d\right\}$. Equation (5) implies $\max \left\{{\nu }_2(b),{\nu }_2(c)\right\}=1$. Equation (2) implies $\max \left\{{\nu }_2(a),{\nu }_2(b)\right\}=3$. Equation (6) implies $\max \left\{{\nu }_2(b),{\nu }_2(d)\right\}=2$. Equation (1) implies ${\nu }_2(a)+{\nu }_2(b)+{\nu }_2(c)+{\nu }_2(d)=6$. Therefore, all above jointly imply ${\nu }_2(a)=3$, ${\nu }_2(d)=2$, and $\left({\nu }_2(b),{\nu }_2(c)\right)=\left(0,1\right)$ or $\left(1,0\right)$. Second, we compute ${\nu }_3(x)$ for $x\in \left\{a,b,c,d\right\}$. Equation (2) implies $\max \left\{{\nu }_3(a),{\nu }_3(b)\right\}=2$. Equation (3) implies $\max \left\{{\nu }_3(a),{\nu }_3(c)\right\}=3$. Equation (4) implies $\max \left\{{\nu }_3(a),{\nu }_3(d)\right\}=3$. Equation (1) implies ${\nu }_3(a)+{\nu }_3(b)+{\nu }_3(c)+{\nu }_3(d)=9$. Therefore, all above jointly imply ${\nu }_3(c)=3$, ${\nu }_3(d)=3$, and $\left({\nu }_3(a),{\nu }_3(b)\right)=\left(1,2\right)$ or $\left(2,1\right)$. Third, we compute ${\nu }_5(x)$ for $x\in \left\{a,b,c,d\right\}$. Equation (5) implies $\max \left\{{\nu }_5(b),{\nu }_5(c)\right\}=2$. Equation (2) implies $\max \left\{{\nu }_5(a),{\nu }_5(b)\right\}=3$. Thus, ${\nu }_5(a)=3$. From Equations (5)-(7), we have either ${\nu }_5(b)\le 1$ and ${\nu }_5(c)={\nu }_5(d)=2$, or ${\nu }_5(b)=2$ and $\max \left\{{\nu }_5(c),{\nu }_5(d)\right\}=2$. Equation (1) implies ${\nu }_5(a)+{\nu }_5(b)+{\nu }_5(c)+{\nu }_5(d)=7$. Thus, for ${\nu }_5(b)$, ${\nu }_5(c)$, ${\nu }_5(d)$, there must be two 2s and one 0. Therefore, $$\begin{array}{rl}\gcd (a,b,c,d)& ={\Pi }_{p\in \{2,3,5\}}{p}^{\min \{{\nu }_p(a),{\nu }_p(b),{\nu }_p(c),{\nu }_p(d)\}}\\ & =2^0\cdot 3^1\cdot 5^0\\ & =\boxed{3}.\end{array}$$