2016 AMC 12B Problem 24

Writing each entry as $77$ times a reduced value, we need $\gcd=1$ and $\text{lcm}=m=\frac{n}{77}.$ For each prime $p$ dividing $m$ with maximum exponent $M,$ the number of valid exponent quadruples is $(M+1)^4-2M^4$ $+(M-1)^4$ $=2(6M^2+1).$ The total over all primes must equal $77{,}000=2^3\cdot5^3\cdot7\cdot11.$ Since $2(6M^2+1)$ equals $14,$ $50,$ and $110$ for $M=1,2,3,$ and $14\cdot50\cdot110=77{,}000,$ the exponents $1,2,3$ give one candidate factorization. Every prime contributes exactly one factor of $2,$ so exactly three primes divide $m.$ Their odd factors $6M^2+1$ must divide $\frac{77{,}000}{2^3}=9625.$ Checking the divisors gives $M=1,2,3,8,$ with odd factors $7,25,55,385.$ The choice $M=8$ leaves only $25$ for the product of the other two odd factors, but each is at least $7,$ so this is impossible. Therefore the maximum exponents are exactly $1,2,3.$ To minimize $m=\frac{n}{77},$ assign the largest exponent to the smallest prime: $m=2^3\cdot3^2\cdot5=360,$ so $n=77\cdot360=27{,}720.$ Thus, the correct answer is D.