Cubic of an Integer

Let $x$ be an integer chosen uniformly between $1$ and $10^{12}$ . What is the probability that $x^3$ ends in the digits $11$ ?

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#Only the last two digits matter

The last two digits of $x^3$ depend only on the last two digits of $x$ , since every higher place of $x$ adds a multiple of $100$ to the cube. So the question is really about $x \bmod 100$ , and I can pin it down one digit at a time.

#The units digit

A cube ends in $1$ only when its base does. Running the units digit $d$ from $0$ to $9$ , the cubes end in $0, 1, 8, 7, 4, 5, 6, 3, 2, 9$ , and $1$ appears exactly once, at $d = 1$ . So $x$ ends in $1$ .

#The tens digit

Write $x = 10a + 1$ , where $a$ collects the digits above the units. Then

x^3 = (10a + 1)^3 = 1000a^3 + 300a^2 + 30a + 1 \equiv 30a + 1 \pmod{100}, \tag{1}

because the first two terms are multiples of $100$ . Ending in $11$ asks for $30a + 1 \equiv 11$ , that is $3a \equiv 1 \pmod{10}$ . Since $3$ is invertible modulo $10$ , the unique solution is $a \equiv 7$ , so $x \equiv 71 \pmod{100}$ . A quick check gives $71^3 = 357911$ .

1..01

11..31

21..61

31..91

41..21

51..51

61..81

71..11

81..41

91..71

Only integers ending in 1 can cube to a number ending in 1. Among the ten such residues modulo 100, the cube's last two digits run through every value of the form 30a + 1, and exactly one of them, 71, lands on 11.

#The probability

Exactly one residue in every hundred consecutive integers cubes to a number ending in $11$ . The range $1$ to $10^{12}$ holds $10^{12} / 100 = 10^{10}$ of them, so the probability is

\frac{10^{10}}{10^{12}} = \frac{1}{100}. \tag{2}

Nothing here leans on the value $10^{12}$ beyond its being a multiple of $100$ . The density of integers whose cube ends in $11$ is exactly $1/100$ .