\( k \) is divisible by the sum of its digits.

February 22, 2026

$\( k \) is divisible by the sum of its digits.$

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\( n \equiv 2 \pmod{3}, n \equiv 1 \pmod{5} \): Let \( n = 5k + 1 \). Then \( 5k + 1 \equiv 2 \pmod{3} \Rightarrow 2k + 1 \equiv 2 \Rightarrow 2k \equiv 1 \Rightarrow k \equiv 2 \pmod{3} \Rightarrow k = 3m + 2 \Rightarrow n = 5(3m+2)+1 = 15m + 11 \). Smallest is \( n = 11 \).
\( n \equiv 2 \pmod{3}, n \equiv 4 \pmod{5} \): \( n = 5k + 4 \). Then \( 5k + 4 \equiv 2 \pmod{3} \Rightarrow 2k + 1 \equiv 2 \Rightarrow 2k \equiv 1 \Rightarrow k \equiv 2 \pmod{3} \Rightarrow k = 3m + 2 \Rightarrow n = 15m + 14 \). Smallest \( n = 14 \).
\( k \equiv 2 \pmod{7} \),
\( k = 16 \): digit sum = \( 1+6 = 7 \), \( 16 \div 7 \approx 2.28 \), not divisible.
\( k = 23 \): sum = \( 2+3 = 5 \), \( 23 \div 5 = 4.6 \), no.
\( k = 30 \): sum = \( 3+0 = 3 \), \( 30 \div 3 = 10 \), yes!

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