2017 AMC 10A Problem 25

Whether a number has the property depends only on which digits it uses, not their order. So first find which sets of digits work, then count how many three-digit numbers each set makes.

Step 1: the test for $11$.

1. A three-digit number with digits $p, q, r$ (in that order) is a multiple of $11$ exactly when $p - q + r$ is a multiple of $11$. For digits, $p - q + r$ is between $-9$ and $18$, so it must be $0$ or $11$.
- $p - q + r = 0$ means the middle digit is the sum of the outer two. Example: $121$, since $1 + 1 = 2$.
- $p - q + r = 11$ means the middle digit is the sum of the outer two minus $11$. Example: $209$, since $2 + 9 - 11 = 0$.
2. So a set of three digits works when one digit equals the sum of the other two, or that sum minus $11$. (Put that digit in the middle. You can always choose the order of the outer two so the number does not start with $0$.)

Step 2: count sets and their arrangements. Split by whether a digit repeats and whether $0$ appears, because that changes how many arrangements there are.

3. Three different nonzero digits. Each such set can be written in $6$ orders.
- One digit is the sum of the other two, $u + v$ with $u < v$. For sums $3, 4, 5, 6, 7, 8, 9$ the pairs number $1, 1, 2, 2, 3, 3, 4$ (for example, sum $7$: $1{+}6, 2{+}5, 3{+}4$). That is $16$ sets.
- One digit is $u + v - 11$. The sum must be $12$ to $17$ for this to be a nonzero digit. The pairs number $3, 3, 2, 2, 1, 1$ (for example, sum $12$: $3{+}9, 4{+}8, 5{+}7$, each with third digit $1$). That is $12$ sets.
- No set is in both lists. Total $28$ sets, giving $28 \times 6 = 168$ numbers.
4. A repeated nonzero digit, like $d, d, e$. Each set can be written in $3$ orders.
- $e = d + d$: $d = 1, 2, 3, 4$ (sets $112, 224, 336, 448$).
- $e = d + d - 11$: $d = 6, 7, 8, 9$ (third digit $1, 3, 5, 7$).
- Total $8$ sets, giving $8 \times 3 = 24$ numbers.
5. A zero and two different nonzero digits $0, u, v$. The only way to fit the rule is $u + v - 11 = 0$, so $u + v = 11$: the pairs $2{+}9, 3{+}8, 4{+}7, 5{+}6$. That is $4$ sets. Each makes $4$ three-digit numbers (six orders, minus the two that start with $0$): $4 \times 4 = 16$ numbers.
6. A zero and a repeated digit $0, d, d$. Always works, because $d = d + 0$ (the number $d\,d\,0$ has $d - d + 0 = 0$). There are $9$ sets ($d = 1$ to $9$), each making $2$ numbers ($dd0$ and $d0d$): $9 \times 2 = 18$ numbers.
7. Add: $168 + 24 + 16 + 18 = 226$.

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

When a question is about rearranging digits, the real object is the set of digits, so count those sets first and arrangements second. That avoids counting a number twice when its digits can form two different multiples of $11$. The "first minus middle plus last" test for $11$ turns the question into a simple rule about sums of digits.