How many integers between and , inclusive, have the property that some permutation of its digits is a multiple of between and For example, both and have this property:
Pick one
Solution
There are 81 multiples of 11 between and inclusive. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Switching shows we have overcounted by a factor of 2, so assign permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have as a digit. Since cannot be the digit of the hundreds place, we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243) to get 226.
If short on time, observe that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
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