A number from 1 to 1000 is selected. What is P(The Last Two Digits of the Cube = 1)? Y.A. Rozanov. Probability Theory: A Concise Course Chapter 1 Problem 9.  
A number from 1 to 1000 is selected at random. What is probability that the last two digits of  it's cube are equal to 1?
The book reports that the  answer is .01.
I believe the  answer follows from the fact that number $100x^2+10y+z$ cubed has the form $(100x^2+10y+z)^3$ = A trinomial expansion and using this knowledge to somehow show that there are only two solutions and each is 1/10 likely. 
 A: HINT: Suppose that the number is $n=100a+10b+c$, where $a,b$, and $c$ are decimal digits. Let $m=10a+b$. Then
$$\begin{align*}
n^3&=(10m+c)^3\\
&=1000m^3+300m^2c+30mc^2+c^3\\
&=100\left(10m^3+3m^2c\right)+30mc^2+c^3\;.
\end{align*}$$
The first term in the last line clearly has no effect on the last two digits of $n^3$, and the second has no effect on the last digit. The last digit of $n^3$ is the last digit of $c^3$, which is $1$ if and only if $c=1$. You’d pretty much arrived at this point on your own.
In that case the second-last digit of $n^3$ is the second-last digit of $30mc^2$, which is the last digit of $3m$, and that’s the last digit of $3b$. When is the last digit of $3b$ equal to $1$?
A: Hint: The number $3$ has no common divisors with $\phi(100)=|\mathbb{Z}_{100}^*|=40$, so cubing is a permutation of the group $\mathbb{Z}_{100}^*$. If a number has common divisor with $100$ its cube ain't gonna end with two ones. 
No need to work out the possibilities for $n$ :-)
The upshot here is that this argument goes through unchanged, if we replace the pair of digits $11$ with any other pair such that the latte digit is one of $1,3,7,9$. Hence the probability is $1/100$ for all such pairs.
