余数问题

设所求最小正整数为 $X$。题意等价于

$$X = 2X_2 + 1 \\ X = 3X_3 + 2 \\ \vdots \\ X = 10X_{10} + 9$$

令 $X'_i = X_i + 1$，则两边同时加 1 得到

$$X+1 = 2X'_2 \\ X+1 = 3X'_3 \\ \vdots \\ X+1 = 10X'_{10}$$

也就是说，$X+1$ 必须能被 $2,3,4,5,6,7,8,9,10$ 同时整除。这些数的最小公倍数是 2520，因此

$$X+1=2520 \quad\Rightarrow\quad X=\boxed{2519}$$

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Original Explanation

Let $X$ denote the smallest positive integer we are solving for. We can further define $X_2, X_3, ..., X_{10}$ as positive integers such that:

$$X = 2 \times X_2 + 1 \\ X = 3 \times X_3 + 2 \\ \vdots \\ X = 10 \times X_{10} + 10$$

Let $X'_i = X_i + 1 \forall i \in \{2, 3, ..., 10\}$.

Then, adding $1$ to both sides of each equations:

$$X + 1 = 2 \times X'_2 \\ X + 1 = 3 \times X'_3 \\ \vdots \\ X + 1 = 10 \times X'_{10}$$

In other words, we are now looking for $X$ such that $X+1$ is perfectly divisible by 2, 3, 4, 5, 6, 7, 8, 9, and 10. The least common multiple of these numbers is 2520, and thus $X$ is $\boxed{2519}$