68(1178). What digit does the number end with: a) $2^{1000}$; b) $3^{1000}$; c) $7^{1000}$?
Solution. Method I. a) Let's determine which digit numbers of the form $2^{n}$ end with when $n=1,2,3, \ldots$. We have: $2^{1}$ ends with the digit $2, 2^{2}$ ends with the digit $4, 2^{3}$ with the digit $8, 2^{4}$ with the digit $6, 2^{5}$ with the digit 2 (just like $2^{1}$). Therefore, the digits $2,4,8,6$ at the end of numbers of the form $2^{n}$ will repeat every four powers. Since $2^{1000}=2^{4 \cdot 250}$, then $2^{1000}$, like $2^{4}$, ends with the digit 6.

b) The powers $3^{1}, 3^{2}, 3^{3}, 3^{4}, 3^{5}$ end with the digits $3,9,7,1,3$. Therefore, the number $3^{1000}$, like the number $3^{4}$, ends with the digit 1 (since the digits at the end of numbers of the form $3^{n}$ repeat every four powers, and $1000=4 \cdot 250$).

c) Similarly, we establish that the digits $7,9,3,1$ at the end of numbers of the form $7^{n}$ repeat every four powers. Therefore, $7^{1000}$, like $7^{4}$, ends with the digit 1.

Method II. a) Since $2^{1000}=\left(2^{250}\right)^{4}$, and the number $2^{250}$ is an even number not divisible by 10 (ends with a digit other than zero), its fourth power ends with the digit 6. (We use the result obtained in solving problem 67.)

b) Since $3^{4}=81, 3^{1000}=\left(3^{4}\right)^{250}=81^{250}$, and a number ending with the digit 1, when raised to any natural power, will give a number ending with 1, then $3^{1000}$ ends with the digit 1.

c) Since $7^{1000}=\left(7^{4}\right)^{250}$, and the number $7^{4}$ ends with the digit 1, then the number $7^{1000}$ will also end with the digit 1.