69*. Determine the five last digits of the number

$$
\left.N=9^{\left(9^{(} \cdot 9^{\left(9^{9}\right)}\right)} \ldots\right),
$$

written using 1001 nines in a similar manner to the number in problem 68).
69. Let's consider the sequence of numbers:

$$
\begin{aligned}
& 1^{\circ} \cdot Z_{1}=9 \\
& \begin{aligned}
2^{\circ} \cdot Z_{2} & =9^{Z_{1}}=(10-1)^{Z_{1}}= \\
& =10^{Z_{1}}-C_{Z_{1}}^{1} \cdot 10^{Z_{1}-1}+\ldots+C_{Z_{1}}^{1} \cdot 10-1
\end{aligned}
\end{aligned}
$$

where the omitted terms in the expansion are all divisible by 100. But \( C_{Z_{1}}^{1}=9 \); therefore, the last two digits of the number \( Z_{2} \) will be the same as the last two digits of the number \( 9 \cdot 10-1=89 \).

$$
\text { 3. } \begin{aligned}
Z_{3}=9^{Z_{2}} & =(10-1)^{Z_{2}}= \\
& =10^{Z_{2}}-C_{Z_{2}}^{1} \cdot 10^{Z_{1}-1}+\ldots-C_{Z_{3}}^{2} \cdot 10^{2}+C_{Z_{2}}^{1} \cdot 10-1 .
\end{aligned}
$$

But \( Z_{2} \) ends in 89; therefore, \( C_{Z_{2}}^{1}=Z_{2} \) ends in 89, and \( C_{Z_{2}}^{2}=\frac{Z_{2}\left(Z_{2}-1\right)}{1 \cdot 2}=\frac{\ldots 89 \cdot \ldots 88}{1 \cdot 2} \) (dots represent unknown digits) ends in 6. Therefore, the last three digits of the number \( Z_{3} \) will be the same as the last three digits of the number \( -600+890-1=289 \).

$$
\begin{aligned}
4^{7} & : Z_{4}=9^{Z_{3}}=(10-1)^{Z_{3}}= \\
& =10^{Z_{3}}-C_{Z_{3}}^{1} \cdot 10^{Z_{3}-1}+\ldots \div C_{Z_{1}}^{3} \cdot 10^{3}-C_{Z_{3}}^{2} \cdot 10^{2}+C_{Z_{3}}^{1} \cdot 10-1
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_b937612abb2773f7abffg-126.jpg?height=54&width=969&top_left_y=671&top_left_x=52)

$$
C_{Z_{3}}^{2}=\frac{Z_{3}\left(Z_{3}-1\right)}{1 \cdot 2}=\frac{\ldots 289 \cdot \ldots 288}{1 \cdot 2}
$$

ends in 16;

$$
C_{Z_{3}}^{3}=\frac{Z_{3}\left(Z_{3}-1\right)\left(Z_{3}-2\right)}{1 \cdot 2 \cdot 3}=\frac{\ldots 289 \ldots 288 \ldots 287}{1 \cdot 2 \cdot 3}
$$

ends in 4. Therefore, the last four digits of the number \( Z_{4} \) will be the same as the last four digits of the number \( 4000-1600+2890-1=5289 \).

$$
\begin{aligned}
5^{0} \cdot Z_{5}=9^{Z_{k}}= & (10-1)^{Z_{k}}=10^{Z_{4}}-C_{Z_{1}}^{1} \cdot 10^{Z_{4}-1}+\ldots \\
& \ldots-C_{Z_{4}}^{1} \cdot 10^{1}+C_{Z_{4}}^{3} \cdot 10^{3}-C_{Z_{4}}^{2} \cdot 10^{2}+C_{Z_{k}}^{1} \cdot 10-1
\end{aligned}
$$

Since \( Z_{4} \) ends in 5289, \( C_{Z_{4}}^{1}=Z_{4} \) ends in 5289;

$$
\cdot C_{Z_{4}}^{\prime 2}=\frac{Z_{4}\left(Z_{1}-1\right)}{1 \cdot 2}=\frac{\ldots 5289 \cdot \ldots 5288}{1 \cdot 2}
$$

ends in 116;

$$
C_{Z_{4}}^{3}=\frac{Z_{4}\left(Z_{4}-1\right)\left(Z_{4}-2\right)}{1 \cdot 2 \cdot 3}=\frac{\ldots 5289 \cdot \ldots .5288 \cdot \ldots 5287}{1 \cdot 2 \cdot 3}
$$

ends in 64; finally,

$$
C_{Z_{4}}^{4}=\frac{Z_{4}\left(Z_{4}-1\right)\left(Z_{4}-2\right)\left(Z_{4}-3\right)}{1 \cdot 2 \cdot 3 \cdot 4}=\frac{\ldots 5289 \ldots 5288 \cdot \ldots 5287 \ldots 5286}{1 \cdot 2 \cdot 3 \cdot 4}
$$

ends in 6. Therefore, \( Z_{5} \) ends in the same last five digits as the number

$$
-60000+64000-11600+52890-1=45289
$$

From the fact that the last four digits of the number \( Z_{5} \) coincide with the last four digits of the number \( Z_{4} \), it follows that the last five digits of the number \( Z_{6}=9^{Z_{6}}=(10-1)^{Z_{\overline{5}}} \) coincide with the last five digits of the number \( Z_{5} \) ( \( \left.=9^{Z_{4}}\right) \). Similarly, it can be shown that all numbers in the sequence

$$
Z_{3}, \quad Z_{6}=9^{Z_{5}}, Z_{7}=9^{Z_{5}}, \ldots, Z_{1000}=9^{Z_{289}}, Z_{1001}=9^{Z_{1000}}
$$

end in the same five digits, namely 45289. But \( Z_{1001} \) is the number \( N \) in the problem condition.