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  "body": "\n## Statement\n\nLet $n$ be a natural number with decimal digits $d_k d_{k-1} \\dots d_0$, so that $n = \\sum_{i=0}^{k} d_i 10^i$.\nThen $11 \\mid n$ if and only if $11 \\mid d_0 - d_1 + d_2 - \\dots + (-1)^k d_k$.\n\n## Remarks\n\nMore precisely, $n \\equiv \\sum_{i=0}^{k} (-1)^i d_i \\pmod{11}$. This holds because $10 \\equiv -1 \\pmod{11}$, so $10^i \\equiv (-1)^i \\pmod{11}$.\nFor example, $918{,}082$ gives $2 - 8 + 0 - 8 + 1 - 9 = -22$, and $11 \\mid -22$. So $11 \\mid 918{,}082$.\n\n## Explanations\n\n### L1 human\n\nStart at the last digit, and add and subtract the digits in turn.\nThe number is a multiple of $11$ exactly when the result is a multiple of $11$, including $0$.\nFor example, $121$ gives $1 - 2 + 1 = 0$, and indeed $121 = 11 \\cdot 11$.\n",
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