How can "casting out nines" be used to find S(S(S(n))), where n=7777^7777 and S(n)=the sum of the digits of n

Casting out nines is also the theorem that states if k=a (mod 9), then d1+d2+...+dn=a (mod 9), where k=d1d2d3...dn (the normal decimal representation of k).

Well... if you mean to take S(S(S...(n)) enough times to get a number between 0 and 8, then the answer is:


7777^7777 mod 9 === (7777 mod 9)^7777 === 1^7777 === 1 mod 9

So eventually, the answer will be 1.