2023 AMC 12B Problem 24

Handle each prime separately using the exponents of $a,b,c,d.$ Prime $2$ (total $6$): $\max(b,c)=1$ forces $a=3;$ then $b+c+d=3$ with $\max(b,d)=\max(c,d)=2$ gives $d=2$ and $\{b,c\}=\{0,1\},$ so the minimum exponent is $0.$ Prime $3$ (total $9$): $\max(a,b)=2$ with the other lcms equal to $3$ forces $c=d=3;$ then $a+b=3$ with $\max(a,b)=2$ gives $\{a,b\}=\{1,2\},$ so the minimum is $1.$ Prime $5$ (total $7$): $\max(a,b)=3$ with $\max(b,c),$ $\max(b,d),\max(c,d)=2$ forces $a=3;$ then $b+c+d=4$ with each $\le 2$ and pairwise maxima $2$ gives two of them equal to $2$ and one equal to $0,$ so the minimum is $0.$ Therefore $\gcd(a,b,c,d)=2^0\cdot 3^1\cdot 5^0=3.$ Thus, the correct answer is C.