Problem:
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.
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Solution:
Recall the divisibility test for : A three-digit number is divisible by if and only if is divisible by . The smallest and largest three-digit multiples of are, respectively, and , so the number of three-digit multiples of is . They may be grouped as follows:
There are multiples of that have the form for . They can each be permuted to form a total of three-digit integers. In each case is a multiple of and is not, so these multiples of give integers with the required property.
There are multiples of that have the form , namely , , and . They can each be permuted to form a total of three-digit integers. In each case is a multiple of , but neither nor is, so these multiples of give integers with the required property.
If a three-digit multiple of has distinct digits and one digit is , it must have the form with . There are such integers, namely . They can each be permuted to form a total of three-digit integers, but these multiples of give only distinct sets of permutations, leading to integers with the required property.
The remaining three-digit multiples of all have the form , where , and are distinct nonzero digits. They can each be permuted to form a total of three-digit integers, and in each case both and - and only these - are multiples of . Therefore these 56 multiples of give only distinct sets of permutations, leading to integers with the required property.
The total number of integers with the required property is .
The problems on this page are the property of the MAA's American Mathematics Competitions