随机整数立方末两位为 11 的概率

等价于求满足

$$x^3\equiv 11\pmod{100}$$

的 $x\bmod 100$ 的个数。

直接枚举可得唯一解为 $x\equiv 71\pmod{100}$，因此概率为

$$\frac{1}{100}=0.01.$$

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Original Explanation

Write $x = a + 10b$. Then

$$x^3 = (\,a + 10b\,)^3 = a^3 \;+\; 30\,a^2\,b \;+\; 300\,a\,b^2 \;+\; 1000\,b^3.$$

For the last digit of $x^3$ to be 1, we need $a^3 \equiv 1 \pmod{10}$, so $a \equiv 1 \pmod{10}$. Hence $x$ must end in 1. For the second-to-last digit also to be 1, more analysis shows $x$ must end in 71. Among the integers from 1 to $10^{12}$, exactly 1% end with 71. Thus, the probability is 0.01.