Nine is the key

Proposition.- ${n} \equiv {f(n)} \pmod{9}$ for each n natural number

Proof.- Let

$n= abcde = e + 10d + 100c + 1000b + 10000a$

$\Rightarrow f(n) = a + b + c +d + e \Rightarrow 9 |$

$n - f(n) = 9d + 99c + 999b + 9999a.$

This reasoning can be generealized...Q.E.D

So $5^{10000} \equiv {f(5^{10000})} \pmod{9}$ .

$5^{10000}$ = $(5^{4})^{2500} \equiv {4^{2500}} \pmod9$ $\Rightarrow$ $4^{2500}$ = ${4 \cdot (4^{3})^{833}} \equiv {4 \cdot 1^{833}} \equiv 4 \pmod{9}$ $\Rightarrow$ $5^{10000} \equiv 4 \pmod{9}$ .

Since $5^{10000}$ has at most 9999 digits and $9 \cdot 9999$ =89991, 1< f( $5^{10000}$ ) < 89991 < 99999, so f( $5^{10000}$ ) has at most 5 digits $\Rightarrow$ 1 < f(f( $5^{10000}$ )) < 45, ( f(f( $5^{10000}$ )) $\equiv$ 4 (mod 9) ) $\Rightarrow$ f(f( $5^{10000}$ )) = 4, 13, 22, 31 or 40( They add all the same) $\Rightarrow$ f(f(f( $5^{10000}$ ))) = 4

According to the question we need to find sum of $5^n$ three times. Let us observe the pattern of sum three times :

$5^1 = 5$

$5^2 = 7$

$5^3 = 8$

$5^4 = 4$

$5^5 = 2$

$5^6 = 1$

$5^7 = 5$

we found that its cyclicity is 6,

so answer is 4

See problem 4 of the 1975 IMO .

The solution can be imitated

Solving f(5^1000) with C++