We are given a uniform random integer \( x \) from \( 1 \) to \( 10^{12} \), and are asked to compute the probability that the decimal representation of \( x^3 \) ends in the digits “11”.

This is equivalent to determining the number of integers \( x \in {1, 2, \dots, 10^{12}} \) for which

\[ x^3 \equiv 11 \pmod{100} \]

and dividing that count by \( 10^{12} \).

Modular Reduction Strategy

We aim to solve:

\[ x^3 \equiv 11 \pmod{100} \]

Since \( 100 = 4 \times 25 \), we solve this congruence modulo 4 and modulo 25 separately, and then combine the solutions using the Chinese Remainder Theorem (CRT).

Step 1: Modulo 4

We examine all residues \( x \mod 4 \):

  • \( 0^3 \equiv 0 \mod 4 \)
  • \( 1^3 \equiv 1 \mod 4 \)
  • \( 2^3 \equiv 8 \equiv 0 \mod 4 \)
  • \( 3^3 \equiv 27 \equiv 3 \mod 4 \)

So the possible cube residues mod 4 are \( {0, 1, 3} \). Thus, \( x^3 \equiv 11 \equiv 3 \mod 4 \) implies:

\[ x \equiv 3 \mod 4 \]

Step 2: Modulo 25

We need:

\[ x^3 \equiv 11 \mod 25 \]

Try small values:

  • \( 1^3 = 1 \)
  • \( 2^3 = 8 \)
  • \( 3^3 = 27 \)
  • …
  • \( 16^3 = 4096 \equiv 4096 \mod 25 = 4096 - 25 \times 163 = 4096 - 4075 = 21 \)
  • \( 17^3 = 4913 \equiv 4913 - 25 \times 196 = 4913 - 4900 = 13 \)
  • \( 18^3 = 5832 \equiv 5832 - 5825 = 7 \)
  • \( 19^3 = 6859 \equiv 6859 - 6825 = 34 \mod 25 = 9 \)
  • \( 20^3 = 8000 \equiv 8000 \mod 25 = 0 \)
  • \( 21^3 = 9261 \equiv 9261 - 9250 = 11 \)

Success! \( x \equiv 21 \mod 25 \) is a solution.

Verify uniqueness: Since the cube map modulo 25 is not injective, we check if there are other solutions.

Try:

  • \( x = 71 \): then \( x = 25 \cdot 2 + 21 = 71 \)
  • \( x^3 = 71^3 = (70 + 1)^3 = 343000 + 3 \cdot 4900 + 3 \cdot 70 + 1 = 357911 \)
  • Ends in 11 ✔️

So \( x \equiv 21 \mod 25 \) yields \( x^3 \equiv 11 \mod 100 \) if and only if \( x \equiv 3 \mod 4 \)

Use CRT to combine:

Find \( x \mod 100 \) such that:

  • \( x \equiv 3 \mod 4 \)
  • \( x \equiv 21 \mod 25 \)

Let \( x = 25k + 21 \). Plug into first congruence:

\[ 25k + 21 \equiv 3 \mod 4 \Rightarrow k \equiv 2 \mod 4 \]

So \( k = 4m + 2 \), thus:

\[ x = 25k + 21 = 25(4m + 2) + 21 = 100m + 71 \]

Therefore, all solutions are:

\[ x \equiv 71 \mod 100 \]

Final Probability

Among \( 1 \) to \( 10^{12} \), every 100th number satisfies \( x \equiv 71 \mod 100 \). Hence, there are:

\[ \left\lfloor \frac{10^{12}}{100} \right\rfloor = 10^{10} \]

such integers, and the probability is:

\[ \frac{10^{10}}{10^{12}} = \boxed{\frac{1}{100}} \]

Reference