Sum of the digits of $N=5^{2012}$ The sum of the digits of $N=5^{2012}$ is computed.
The sum of the digits of the resulting sum is then computed.
The process of computing the sum is repeated until a single digit number is obtained.
What is this single digit number?
 A: You want to know the value of $5^{2012} \pmod 9$.
Since $\varphi(9) = 3^2 - 3 = 6$ and $\gcd(5,9) = 1$, then, by Euler's theorem,  $5^6 \equiv 1 \pmod 9$.
Since $2012 = 335 \times 6 + 2$, 
$$5^{2012} \equiv (5^6)^{335} \times 5^2 
\equiv 1^{335} \times 25 \equiv 7 \pmod 9.$$
A: If Euler's Theorem is unknown then we can instead use the Binomial Theorem. By casting nines the iterated digit sum  equals the remainder mod $9$, which we may compute mentally as follows
${\rm mod}\ \color{#c00}9\!:\,\ 5^n\equiv (-4)^n \equiv (-1)^n (1+3)^n \equiv (-1)^n(1 + 3n+ \color{#c00}{3^2}(\cdots))\equiv (-1)^n(1+3n)$
therefore $\ n = 2012\,\Rightarrow\,5^n \equiv 1+3(2012)\equiv 1+3\,\underbrace{(2\!+\!0\!+\!1\!+\!2)}_{\large \rm cast\ nines}\equiv 16\equiv 7\,\pmod 9$
A: Hint : You can prove that if $d(n)$ denotes sum of digits of $n$, then $n$ is congruent to $d(n)$ mod 9. Also, $N<10^{2012}$ implies $d(N)<9*2012<20000$.
A: Hint: if $s$ is the sum of digits of a number $a$ then
$$s\equiv a\pmod 9.$$
Now all you need too do is compute $5^{2012} \pmod 9$.
A: Let $f(n)$ denote the sum of sum of ... sum of digits of $5^n$, then:
$f(n)=\cases
 {
  1 & $n\equiv0\pmod6$\\
  5 & $n\equiv1\pmod6$\\
  7 & $n\equiv2\pmod6$\\
  8 & $n\equiv3\pmod6$\\
  4 & $n\equiv4\pmod6$\\
  2 & $n\equiv5\pmod6$\\
 }
$
Therefore: $2012\equiv2\pmod6 \implies f(2012)=7$.
