Problem 783

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.