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  "body": "\n## Proof\n\n1. Let $n = \\sum_{i=0}^{k} d_i 10^i$ with decimal digits $d_0, \\dots, d_k$. Let $A = \\sum_{i=0}^{k} (-1)^i d_i = d_0 - d_1 + d_2 - \\dots + (-1)^k d_k$ be the alternating digit sum.\n2. Since $10 - (-1) = 11$, the number $11$ [[divides|096PW7Q014]] $10 - (-1)$. So $10 \\equiv -1 \\pmod{11}$ by the definition of [[congruence|0TPW25CBN6]].\n3. Let $0 \\le i \\le k$. By step 2 and [[compatibility with powers|0J5VGJYYC3]], $10^i \\equiv (-1)^i \\pmod{11}$.\n4. Since $d_i \\equiv d_i \\pmod{11}$, step 3 and [[compatibility with multiplication|072D5G4SK6]] give $d_i 10^i \\equiv (-1)^i d_i \\pmod{11}$.\n5. Apply [[compatibility with addition|00FWBN7F0Y]] $k$ times to the $k + 1$ congruences of step 4. This gives $n \\equiv A \\pmod{11}$.\n6. So there is an integer $t$ with $n - A = 11 t$.\n7. If $11 \\mid A$, say $A = 11 s$, then $n = A + 11 t = 11 (s + t)$, so $11 \\mid n$.\n8. If $11 \\mid n$, say $n = 11 m$, then $A = n - 11 t = 11 (m - t)$, so $11 \\mid A$.\n9. By steps 7 and 8, $11 \\mid n$ if and only if $11 \\mid d_0 - d_1 + d_2 - \\dots + (-1)^k d_k$.\n\n## Idea\n\nSince $10 = 11 - 1$, the number $10$ acts like $-1$ modulo $11$. So the powers of $10$ act like $1, -1, 1, -1, \\dots$, and the number is congruent to its alternating digit sum.\n",
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