The answer is $abc$. Notice if $abc$ has a grandson $t$, then $abc = a + a + \dots + a = b + b + \dots + b = c + c + \dots + c \implies abc \mid t$, contradiction. Let $n > abc$ and choose $v \le bc$ such that $av \equiv n \pmod{bc}$ and let $n - av = bcd$ for $d\ge 1$. We claim $av$ is a grandson of $n$, concluding the proof. Let $n = xa + yb + zc$. We can assume $x < v$ as otherwise clearly $av$ is obtainable. Let us also assume $n = xa + yb + z'c + kbc$ with $z' < b$ and $k\ge 1$. Now if $y + ck < cd$, then $xa+yb+zc=n\le (v-1)a+(cd-1)b+(b-1)c = n - a - b + bc - c < n$ by the given condition $a > bc$, contradiction. Otherwise, $y + ck \ge cd$, and we then have $av = n - bcd$ is obtainable, as desired.