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  "body": "\n## Statement\n\nLet $n_1, \\dots, n_k$ be positive integers that are pairwise [[coprime|0GFKP8F8AG]], and let $N = n_1 n_2 \\cdots n_k$.\nLet $a_1, \\dots, a_k$ be integers.\nThen there is an integer $x$ such that $x \\equiv a_i \\pmod{n_i}$ for each $i$.\nTwo such integers are congruent modulo $N$.\n\n## Remarks\n\nIn the language of rings, the map $\\mathbb{Z}/N\\mathbb{Z} \\to \\mathbb{Z}/n_1\\mathbb{Z} \\times \\cdots \\times \\mathbb{Z}/n_k\\mathbb{Z}$, $\\overline{x} \\mapsto (\\overline{x}, \\dots, \\overline{x})$, is a ring isomorphism.\nFor two moduli $m$ and $n$, write $m u + n v = 1$ with [[Bézout's identity|0DQVHMYAGA]]. Then $x = a_2 m u + a_1 n v$ solves $x \\equiv a_1 \\pmod m$ and $x \\equiv a_2 \\pmod n$.\nThe [[Chinese remainder theorem for non-coprime moduli|04SJTX9P4N]] removes the hypothesis of coprimality.\n\n## History\n\nThe Chinese book Sunzi Suanjing, from the 5th century, contains the oldest known example of the problem. Qin Jiushao gave a complete method in 1247.\n\n## Explanations\n\n### L1 human\n\nFind a number that leaves the remainder $2$ when you divide by $3$, the remainder $3$ by $5$, and the remainder $2$ by $7$.\nThe theorem says that such a number exists, and that it is unique up to multiples of $3 \\cdot 5 \\cdot 7 = 105$.\nThe smallest positive answer is $23$.\n",
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