AMC10 2017 A

Answer (A): Recall the divisibility test for 11: A three-digit number $abc$ is divisible by 11 if and only if $a-b+c$ is divisible by 11. The smallest and largest three-digit multiples of 11 are, respectively, $110=10\cdot 11$ and $990=90\cdot 11$, so the number of three-digit multiples of 11 is $90-10+1=81$. They may be grouped as follows: • There are 9 multiples of 11 that have the form $aa0$ for $1\le a\le 9$. They can each be permuted to form a total of 2 three-digit integers. In each case $aa0$ is a multiple of 11 and $a0a$ is not, so these 9 multiples of 11 give 18 integers with the required property. • There are 8 multiples of 11 that have the form $aba$, namely 121, 242, 363, 484, 616, 737, 858, and 979. They can each be permuted to form a total of 3 three-digit integers. In each case $aba$ is a multiple of 11, but neither $aab$ nor $baa$ is, so these 8 multiples of 11 give 24 integers with the required property. • If a three-digit multiple of 11 has distinct digits and one digit is 0, it must have the form $a0c$ with $a+c=11$. There are 8 such integers, namely 209, 308, 407, …, 902. They can each be permuted to form a total of 4 three-digit integers, but these 8 multiples of 11 give only 4 distinct sets of permutations, leading to $4\cdot 4=16$ integers with the required property. • The remaining $81-(9+8+8)=56$ three-digit multiples of 11 all have the form $abc$, where $a$, $b$, and $c$ are distinct nonzero digits. They can each be permuted to form a total of 6 three-digit integers, and in each case both $abc$ and $cba$—and only these—are multiples of 11. Therefore these 56 multiples of 11 give only 28 distinct sets of permutations, leading to $28\cdot 6=168$ integers with the required property. The total number of integers with the required property is $18+24+16+168=226$.