What are the last two digits of $9^{h}$, starting from the number $a=1$,

$$
b=2^{a}, \quad c=3^{b}, \quad d=4^{c}, \quad e=5^{d}, \quad f=6^{e}, \quad g=7^{f}, \quad h=8^{g} ?
$$
Let's examine the last two digits of the powers of nine.

$$
\begin{aligned}
& 9^{0}=\mathbf{0 1}, \quad 9^{1}=\mathbf{0 9}, \quad 9^{2}=\mathbf{8 1}, \quad 9^{3}=\mathbf{2 9}, \quad 9^{4}=\ldots \mathbf{6 1} \\
& 9^{5}=. .49, \quad 9^{6}=. .41, \quad 9^{7}=. .69, \quad 9^{8}=. .21, \quad 9^{9}=. . \mathbf{8 9} \\
& 9^{10}=. . \mathbf{0 1}, \quad 9^{11}=\ldots \mathbf{0 9}, \quad \text { etc. }
\end{aligned}
$$

We see that the last two digits of $9^{10}$ are the same as those of $9^{0}$, and the last two digits of $9^{11}$ are the same as those of $9^{1}$. We conjecture that the last two digits of the powers of 9 form a period of 10. Since the last two digits of a power of 9 can be obtained by multiplying the last two digits of the previous power by 9 and taking the last two digits of the result, if two powers of 9 have the same last two digits, then the next ones will also have the same last two digits. Thus, the last two digits of the 12th and 2nd powers are the same, ... the last two digits of the 10th and 20th powers are the same, and so the 20th power also ends in 01. Therefore, the last two digits of the powers of 9 indeed form a period of 10. According to this, to find the last two digits of $9^{h}$, it is sufficient to know the last digit of $h=8^{\varrho}$, i.e., the remainder when $h$ is divided by 10. Similarly to before, we now examine the last digit of the powers of 8:

$$
8^{1}=\mathbf{8}, \quad 8^{2}=\mathbf{6 4}, \quad 8^{3}=. . \mathbf{2}, \quad 8^{4}=. . \mathbf{6}, \quad 8^{5} . . \mathbf{8}
$$

We see that we again get a periodic sequence, but this time the period length is 4. Thus, to determine the last digit of $8^{g}$, it is sufficient to know the remainder when $g=7^{f}$ is divided by 4. Since $f=6^{e}$ and $e=0$, $f$ will be an even number: $f=2 f^{\prime}$. Thus, $g=7^{2 f^{\prime}}=49 f^{\prime}$, and hence $g-1=49^{f^{\prime}}-1=49^{f^{\prime}}-1^{f^{\prime}}$ is divisible by $(49-1)=48$, and thus also by 4, so $g$ gives a remainder of 1 when divided by 4. From (2), we see that the last digit of $8^{g}$ is the same as the last digit of $8^{1}$. Thus, this digit is 8, so the last digit of $h=8^{g}$ is 8. According to (1), in this case the last two digits of $9^{h}$ are the same as the last two digits of $9^{8}$, which are 21.

Therefore, the last two digits of $9^{h}$ are 21.

Imre Nagy (Szombathely, Lajos Nagy Gymnasium, 2nd grade)

Remark. The problem can be solved without knowing $a, b, c$, and $d$, the result does not change. We only used the fact that $e$ is not zero to show that $f$ is even.