Note that, if we have $n$ numbers as the pile numbers with common $\gcd=d, d>2.$ So we have say $da_1,~da_2,\dots, da_n.$ Then whatever moves we make, the resultant pile numbers will be divisible by $d.$ For example, say we have a pile having $$115~~100~~100 \rightarrow 115,~~50,~~50,~~50,~~50$$Then whatever move we do, we can atmost reach $$ 5,~~5\dots \dots,5~~5.$$ Now, note that $\gcd(81, 115+119=234)=9.$ So the pile number will always be divisible by $9.$ For this case we can't achieve pile numbers as $1,1,\dots 1$ $315$ times.. Now, note that $\gcd(115,81+119=200)=5.$ So the pile number will always be divisible by $5.$ For this case we can't achieve pile numbers as $1,1,\dots 1$ $315$ times.. Now, note that $\gcd(119,81+115=196)=7.$ Same reason. Hence Ah Meng can not divide the marbles into $315$ piles, each with a single marble.