Determine the probability that the last two digits of the…

Determine the probability that the last two digits of the cube of a randomly chosen integer $N$ are ones (i.e. equal to $11$).

Step 1: Only the last two digits of $N$ matter

Write $N = 100q + x$, where $x\in\{0,1,\dots,99\}$ is formed by the last two digits of $N$. Expanding,

\[ N^3 = (100q+x)^3 = 100\cdot(\text{an integer}) + x^3, \]

because every term except $x^3$ contains the factor $100$. So $N^3$ and $x^3$ have the same last two digits.

For a randomly chosen integer each of the $100$ endings $x$ is equally likely, so

\[ P = \frac{\text{number of } x\in\{0,\dots,99\} \text{ with } x^3 \text{ ending in } 11}{100}. \]

Step 2: The last digit

The last digit of $x^3$ depends only on the last digit of $x$:

Every digit appears exactly once in the second row. The cube ends in $1$ only when $x$ ends in $1$. So $x = 10a+1$ with $a\in\{0,1,\dots,9\}$.

Step 3: The tens digit

\[ x^3 = (10a+1)^3 = 1000a^3 + 300a^2 + 30a + 1. \]

The first two terms are multiples of $100$, so the last two digits of $x^3$ are those of $30a+1$. We need $30a+1$ to end in $11$, i.e. $30a$ to end in $10$, i.e. the last digit of $3a$ to be $1$:

Only $a=7$ works ($3\cdot7=21$). So $x = 71$ is the only ending.

Step 4: Check

\[ 71^3 = 357\,911, \qquad 171^3 = 5\,000\,211, \qquad 1071^3 = 1\,228\,480\,911. \]

All end in $11$. ✓

Step 5: The probability

Exactly one of the $100$ equally likely endings is favourable:

\[ \boxed{P = \frac{1}{100} = 0.01.} \]

Step 6: A remark on "randomly chosen integer"

There is no uniform probability on all integers, so the phrase means: choose $N$ at random from a long range such as $1,\dots,100M$; then every ending $00,\dots,99$ occurs exactly $M$ times, and the answer $0.01$ is exact. For other long ranges it is the limiting value.