51. What remainders can the hundredth power of an integer give when divided by 125?
51. Any integer either is divisible by 5 or can be represented in one of the following four forms: $5k+1, 5k+2, 5k-2$, or $5k-1$. If a number is divisible by 5, then its hundredth power is clearly divisible by $5^3 = 125$. Further, using the binomial theorem, we get:

$$
(5k \pm 1)^{100} = (5k)^{100} \pm \ldots + \frac{100 \cdot 99}{1 \cdot 2}(5k)^2 \pm 100 \cdot 5k + 1
$$

where all the omitted terms contain a factor of $5k$ raised to a power of at least 3, and thus are divisible by 125. Similarly,

$$
(5k \pm 2)^{100} = (5k)^{100} \pm \ldots + \frac{100 \cdot 99}{1 \cdot 2}(5k)^2 \cdot 2^{98} \pm 100 \cdot 5k \cdot 2^{99} + 2^{100}
$$

But $\frac{100 \cdot 99}{1 \cdot 2}(5k)^2 = 125 \cdot 990k^2$ and $100 \cdot 5k = 125 \cdot 4k$ are divisible by 125. As for the number $2^{100}$, it can be represented as

$$
(5-1)^{50} = 5^{50} - \ldots + \frac{50 \cdot 49}{1 \cdot 2} \cdot 5 - 50 \cdot 5 + 1
$$

from which it is clear that it gives a remainder of 1 when divided by 125.

Thus, the hundredth power of a number divisible by 5 is divisible by 125; the hundredth power of a number not divisible by 5 gives a remainder of 1 when divided by 125.