10. (20 points) How many zeros does the number $4^{5^{6}}+6^{5^{4}}$ end with in its decimal representation?

#
# Answer: 5

Solution. Consider the number $4^{25}+6$. Let's check that it is divisible by 5 and not divisible by 25. We have

$$
\begin{gathered}
4^{25}+6 \equiv(-1)^{25}+6 \equiv 0 \quad(\bmod 5) \\
4^{25} \equiv 1024^{5}+6 \equiv(-1)^{5}+6 \equiv 5 \quad(\bmod 25)
\end{gathered}
$$

We will write $5^{t} \| c$, if $5^{t} \mid c, 5^{t+1} \nmid c$. For arbitrary integers $a, b$, not divisible by 5, such that $5^{k} \| a-b, k>0$, we will prove that $5^{k+1} \| a^{5}-b^{5}$.

Let $a-b=s, 5^{k} \| s$. We have

$$
a^{5}-b^{5}=s\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)=\quad=s\left((b+s)^{4}+(b+s)^{3} b+(b+s)^{2} b^{2}+(b+s) b^{3}+b^{4}\right)
$$

Expand the brackets using the binomial theorem and consider the expression $\left((b+s)^{4}+(b+s)^{3} b+(b+s)^{2} b^{2}+(b+s) b^{3}+b^{4}\right)$ modulo 25. Clearly, from each bracket only the terms where $s$ is raised to a power no greater than 1 remain. Thus, we have

$$
\begin{aligned}
&(b+s)^{4}+(b+s)^{3} b+(b+s)^{2} b^{2}+(b+s) b^{3}+b^{4} \equiv \\
& \equiv 5 b^{4}+(4+3+2+1) b^{3} s \equiv
\end{aligned}
$$

$$
\equiv 5 b^{4}+10 b^{3} s \equiv 5 b^{4} \quad(\bmod 25)
$$

Thus,

$$
5^{1} \|\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)
$$

Therefore, $5^{k+1} \| a^{2}-b^{2}$.

Now, let's proceed to solving the problem. We have

$$
4^{5^{6}}+6^{5^{4}}=\left(1025^{5}\right)^{5^{4}}-(-6)^{5^{4}}
$$

Let $a=1025^{5}, b=-6$. Apply the proven statement. We get that

$$
5^{2} \|\left(1025^{5}\right)^{5}-(-6)^{5}
$$

Applying the statement three more times, we get that

$$
5^{5} \|\left(1025^{5}\right)^{5^{4}}-(-6)^{5^{4}}
$$

It remains to note that the number from the condition is divisible by $2^{5}$. Thus, the number ends with exactly 5 zeros.

Remark. If we use Hensel's Lemma, the solution becomes obvious.

However, proving the lemma is more complicated than solving this problem.