2017 AMC 10A Problem 25

A three-digit number with digits $p, q, r$ is a multiple of $11$ exactly when $p - q + r \in \{0, 11\}$: the middle digit is the sum of the outer two, or that sum minus $11$. Swapping the outer digits preserves this, so a digit-set is good when one digit equals the sum of the other two, or that sum minus $11$; a good set can always be arranged with a nonzero leading digit. Count good sets by type, then their three-digit arrangements.

Three distinct nonzero digits. One digit is $u + v$ with $u < v$, $u + v \le 9$: sums $3$ through $9$ give $1, 1, 2, 2, 3, 3, 4$ pairs, total $16$. Or one digit is $u + v - 11$ with $u + v \ge 12$: sums $12$ to $17$ give $3, 3, 2, 2, 1, 1$ pairs, total $12$. No set is counted twice (that would force $2u = 11$). $28$ sets, $6$ arrangements each: $168$.

A repeated nonzero digit $\{d, d, e\}$. Need $e = 2d$ ($d = 1, 2, 3, 4$) or $e = 2d - 11$ ($d = 6, 7, 8, 9$): $8$ sets, $3$ arrangements each: $24$.

A zero with two distinct nonzero digits $\{0, u, v\}$. Need $u + v = 11$: $\{2,9\}, \{3,8\}, \{4,7\}, \{5,6\}$. $4$ sets, $4$ arrangements each (zero not in front): $16$.

A zero with a repeated digit $\{0, d, d\}$. Always good since $d = 0 + d$ (e.g. $dd0$): $9$ sets, $2$ arrangements each: $18$.

Total: $168 + 24 + 16 + 18 = 226$.

The answer is $\boxed{\textbf{(A)}\ 226}$.

The property depends only on the multiset of digits, so count multisets first and arrangements second; this sidesteps the double counting that plagues "list all 81 multiples of $11$ and permute" approaches. The divisibility rule for $11$ converts the arithmetic condition into a purely combinatorial one about which digit is a sum of the others, and splitting by repeated digits and zeros keeps the arrangement counts uniform within each case.

• Multiplying the 81 multiples of 11 by the number of permutations, which double counts triples like {2, 0, 9} that give two different multiples (209 and 902).
• Forgetting that permutations with a leading zero are not three-digit numbers, or missing the case $\{0, d, d\}$ that comes from multiples like $550$.