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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 $3 \\mid n$ if and only if $3 \\mid d_0 + d_1 + \\dots + d_k$.\n\n## Remarks\n\nMore precisely, $n \\equiv d_0 + d_1 + \\dots + d_k \\pmod 3$. This holds because $10 \\equiv 1 \\pmod 3$, so $10^i \\equiv 1 \\pmod 3$ by [[compatibility with powers|0J5VGJYYC3]].\nFor example, $3 \\mid 7{,}341$, because $7 + 3 + 4 + 1 = 15$ and $3 \\mid 15$.\nThe rule for [[divisibility by 9|0ESHD7ZQDX]] has the same proof.\n\n## Explanations\n\n### L1 human\n\nAdd the digits of the number. If the sum is a multiple of $3$, the number is a multiple of $3$, and if not, it is not.\nYou can repeat: for $987{,}654$ the sum is $39$, and $3 + 9 = 12$ is a multiple of $3$. So $3 \\mid 987{,}654$.\n",
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