A number from 1 to 1000 is selected. What is P(The Last Two Digits of the Cube = 1)?

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 $1$?

Tags:

Probability

Elementary Number Theory

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