Determine all the numbers formed by three different and non-zero digits, such that the six numbers obtained by
permuting these digits leaves the same remainder after the division by 4.
If ABC is one of the numbers then CBA is another of the numbers. Then ABC-CBA is a multiple of 4.
ABC-CBA = 99*(A-C). Since 99 is coprime to 4 then A-C is a multiple of 4. Since A and C are arbitrary then any pair of A, B, and C differ by a multiple of 4.
Our given pool of digits is {1,2,3,4,5,6,7,8,9} and there is only one set of three digits where all the members differ by multiples of 4: {1,5,9}. Then the set of three digit numbers is {159, 195, 519, 591, 915, 951}.
Solution
