84. a) What is the probability that the square of a randomly chosen integer will end in the digit 1? What is the probability that the cube of a randomly chosen integer will end in the digits 11?

b) What is the probability that the tenth power of a randomly chosen integer ends in the digit 6? What is the probability that the twentieth power of a randomly chosen integer ends in the digit 6?
84. a) The square of an integer ends in the digit 1 if the number itself ends in 1 or 9. This means that among any 10 consecutive integers, exactly two will have squares that end in 1. Therefore, the desired probability is 0.2.

The cube of an integer, as it is easy to verify, ends in the digit 1 only if the number itself ends in 1. The last two digits of the cube of an integer \( n \) obviously depend only on the last two digits of \( n \) (this can be easily seen, for example, from the rule of multiplying multi-digit numbers "in columns"). By calculating the last two digits of the cubes of the numbers 01, 11, 21, 31, 41, 51, 61, 71, 81, and 91, it is easy to verify that among all two-digit numbers, only the cube of 71 ends in 11. Thus, the question of the probability that the cube of a number \( n \) ends in 11 is equivalent to the following: what is the probability that the number \( n \) ends in 71. It is obvious that this probability is 0.01.

b) Obviously, the last digit of the 10th power of an integer \( n \) depends only on the last digit of \( n \). It is easy to verify that \( 4^{10} \) and \( 6^{10} \) end in the digit 6, while the other digits, when raised to the 10th power, result in numbers that do not end in 6 ( \( 2^{10} \) and \( 8^{10} \) end in 4, \( 0^{10} = 0 \), and odd digits, when raised to the 10th power, result in odd numbers). Therefore, out of every 10 consecutive integers, two, when raised to the 10th power, result in numbers ending in the digit 6, while the rest result in numbers that do not end in 6. Thus, the probability that the 10th power of a randomly chosen integer ends in the digit 6 is 0.2.

Since the 10th power of each even number not divisible by 10 ends in the digit 4 or 6, the 20th power of each such number ends in the digit 6. All other integers, when raised to the 20th power, obviously do not result in numbers ending in 6. Therefore, the probability that \( n^{20} \) ends in 6 is 0.4.

Remark. It can even be proven that the probability that the 20th power of an integer ends in 76 is also 0.4 (just as the probability that the 200th power of an integer ends in 376 is 0.4). See the solution to problem 34 in the book by D. O. Shklyarsky, N. N. Chentsov, and I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, Part 1, Moscow, Gostekhizdat, 1954 (Library of Mathematical Circles, Issue 1).