立方数首位概率

末两位只取决于 $x\pmod{100}$，故需解

$$x^3\equiv 51\pmod{100}.$$

等价分解为模 4 与模 25：

• $51\equiv 3\pmod 4$，故 $x\equiv 3\pmod 4$；
• 在模 25 下，逐步可得唯一解 $x\equiv 1\pmod{25}$。

中国剩余定理给出在模 100 下唯一解：

$$x\equiv 51\pmod{100}.$$

在每 100 个连续整数里恰有 1 个满足。又 $10^{12}$ 是 100 的倍数，所以概率为

$$\boxed{\frac{1}{100}}.$$

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英文解析

The last two digits depend only on $x \pmod{100}$, so we need to solve

$$x^3 \equiv 51 \pmod{100}.$$

This is equivalent to decomposing into modulo 4 and modulo 25:

• Since $51 \equiv 3 \pmod 4$, we have $x \equiv 3 \pmod 4$;
• Under modulo 25, we can iteratively obtain the unique solution $x \equiv 1 \pmod{25}$.

The Chinese Remainder Theorem gives the unique solution modulo 100:

$$x \equiv 51 \pmod{100}.$$

There is exactly one such integer in every 100 consecutive integers. Since $10^{12}$ is a multiple of 100, the probability is

$$\boxed{\frac{1}{100}}.$$