Can I receive some assistance please. I was going through some questions when is saw this one,
Let A be the sum of the digits of 16^16. Let B be the sum of the digits of A. What is the sum of the digits of B without finding A.
Well, initially I thought that the sum of the digits of 16^n was (6n + 1) but that fails for n>10. Help?
A little estimating shows that $16^{16}$ has $20$ digits. Therefore $A≤180$. Therefore $B≤27$.
Now, if you kept on iterating the digit sum, you'd get to $7$ since $16^{16}\equiv 7 \pmod 9$. Thus $B\equiv 7 \pmod 9$ So $B\in \{7,16,25\}$.
It follows that the sum of the digits of $B$ is $7$.
Worth noting: a simple search through the possible values for $A$ shows that $25$ is not possible, but both $7$ and $16$ are. I don't see an easy way to eliminate either of these. In truth, $B=16$ but I don't see how to get there without heavier computation.