2017 AMC 10A Problem 25

We can analyze all the multiples of $11$ and see how many permutations each of them contribute. We can do this by casing on the number of unique digits in the number. Case $1:$ all the digits are the same This cannot happen. We can see this by the divisibility rule for $11,$ which says that the sum of the first and last digit minus the middle digit must be divisible by $11.$ If all the digits are the same, then the above expression evaluates to that digit, which cannot be divisible by $11.$ Case $2:$ two of the digits are the same We can split this up into the numbers that have the digit $0$ and those that don’t. There are $8$ multiples of $11$ that do not have the digit $0:$ $$121, 242, 363, 484, 616, 737, 858,$$ and $979.$ Each of these numbers contributes $3$ permutations, so this scenario has $8 \cdot 3 = 24$ numbers. There are $9$ multiples of $11$ that have the digit $0:$$$110, 220, 330, 440, 550, 660, 770,$$ $880,$ and $990.$ For these numbers, $0$ cannot be the hundreds digit, so each of them only contributes $2$ permutations, for a total of $9 \cdot 2 = 18.$ Case $3:$ all the digits are different There are a total of $81$ multiples of $11$ between $100$ and $999.$ The number of these with all different digits is $$81 - 8 - 9 = 64.$$ As in case $2,$ we have to specially account for the numbers with $0$ as a digit. There are $8:$ $$209, 308, 407, 506, 605, 704, 803,$$ and $902.$ Each of these gives us $2 \cdot 2 = 4$ permutations, but we overcount by a factor of $2$ since flipping the first and last digits creates another number already in the set. Therefore, these numbers provide a total of $$8 \cdot 4 \div 2 = 16$$ unique permutations. There are now $64 - 8 = 56$ multiples of $11$ that we need to account for. We know that each of these provides $3! = 6$ permutations. As above, however, note that flipping the first and last digit of any number in this set produces another number in this set. We can see this by using the divisibility rule for $11.$ If $ABC$ is divisible by $11,$ then we have that $A + C - B$ is divisible by $11.$ This means that $C + A - B$ is divisible by $11,$ which means that $CBA$ is also divisible by $11.$ Therefore, these numbers contribute $$56 \cdot 6 \div 2 = 168$$ more permutations. Over all the cases, we have a total of $$24 + 18 + 16 + 168 = 226$$ numbers. Thus, A is the correct answer.