54*. Let $N$ be an even number not divisible by 10. What will be the tens digit of the number $N^{20}$? What will be the hundreds digit of the number $N^{200}$?
54. Let's find the last two digits of the number \(N^{20}\). The number \(N^{20}\) is divisible by 4 (since \(N\) is even). Further, the number \(N\) is not divisible by 5 (otherwise it would be divisible by 10) and, therefore, can be represented in the form \(5k \pm 1\) or in the form \(5k \pm 2\) (see the solution to problem 51). But the number

\[
(5k \pm 1)^{20} = (5k)^{20} \pm 20(5k)^{19} + \ldots + \frac{20 \cdot 19}{1 \cdot 2}(5k)^{2} \pm 20 \cdot 5k + 1
\]

gives a remainder of 1 when divided by 25, and the number

\[
\begin{aligned}
&(5k \pm 2)^{20} = (5k)^{20} \pm 20(5k)^{19} \cdot 2 + \ldots \\
& \ldots + \frac{20 \cdot 19}{1 \cdot 2}(5k)^{2} \cdot 2^{18} \pm 20 \cdot 5k \cdot 2^{19} + 2^{20}
\end{aligned}
\]

gives the same remainder when divided by 25 as the number \(2^{20} = (2^{10})^2 = (1024)^2 = (1025-1)^2\), i.e., also a remainder of 1. From the fact that the number \(N^{20}\) gives a remainder of 1 when divided by 25, it follows that the last two digits of this number can only be 01, 26, 51, or 76. Considering that \(N^{20}\) must be divisible by 4, we get that the last two digits of this number can only be 76. Therefore, the tens digit of the number \(N^{20}\) is 7.

Now let's find the last three digits of the number \(N^{200}\). The number \(N^{200}\) is divisible by 8. Further, since \(N\) is coprime with 5, \(N^{100}\) gives a remainder of 1 when divided by 125 (see the solution to problem 51): \(N^{100} = 125k + 1\). But then \(N^{200} = (125k + 1)^2 = 125^2 k^2 + 250k + 1\) also gives a remainder of 1 when divided by 125. Therefore, \(N^{200}\) can end in the digits 126, 251, 376, 501, 626, 751, or 876; but since \(N^{200}\) is divisible by 8, it must end in the digits 376. Therefore, the hundreds digit of the number \(N^{200}\) is 3.

Note. It is easy to see that the number \(N^{100}\) must already end in the three digits 376.