72*. Find the last 1000 digits of the number

$$
N=1+50+50^{2}+50^{3}+\ldots+50^{999}
$$
72. According to the formula for the sum of the terms of a geometric progression

$$
N=\frac{50^{1000}-1}{50-1}=\frac{80^{1000}-1}{49}
$$

But $\frac{1}{49}$ turns into a pure periodic decimal fraction, the period of which, consisting of 42 digits, is not difficult to find by simple division:

$$
\frac{1}{49}=0.(020408163265306122448979591836734693877551)
$$

or briefly

$$
\frac{1}{49}=0.(P)
$$

where $P$ represents the above sequence of 42 digits.

The nearest integer multiple of 42 to the exponent 1000 is $1008 \approx 24 \cdot 42$. Therefore,

$$
\frac{10^{1008}}{49}=10^{1008} \cdot \frac{1}{49}=\underbrace{P P \ldots P}_{24 \text { times }},
$$

Thus, the fraction

$$
M=\frac{10^{1008}-1}{49}=10^{1008} \cdot \frac{1}{49}-\frac{1}{49}=\underbrace{P P \ldots P}_{24 \text { times }}
$$

is an integer consisting of 1008 digits, which can be divided into 24 groups of repeating 42 digits (in fact, the number $M$ contains not 1008 digits, but 1007, since our number $P$ starts with zero).

Now let's form the difference between the number $N$ we are interested in and the number $M$:

$$
N-M=\frac{5^{1000} \cdot 10^{1000}-1}{49}-\frac{10^{1008}-1}{49}=\frac{5^{1000}-10^{8}}{49} \cdot 10^{10000}
$$

Since the difference $N-M$ of two integers is an integer and $10^{1000}$ is coprime with 49, then $5^{1000}-10^{8}$ must be divisible by 49. Therefore, the number $x=\frac{5^{1000}-10^{8}}{49}$ is an integer and the difference $N-M=10^{1000} \cdot x$ ends with 1000 zeros. Thus, the last 1000 digits of the number $N$ are the same as the last 1000 digits of the number $M$, namely:

$$
p \underbrace{P P \ldots P}_{23 \text { times }}
$$

where $p$ is a group of 34 digits, which are the last 34 digits of the number $P$.