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1 solution
111 = 16 7 - 1 When (16 7 - 1)^333 is expanded by the binomial formula all of the terms will be a multiple of 7 except the last which is (-1)^333 = -1. This shows that 111^333 = 7n - 1 where n is some integer.
333 = 47 7 + 4 When (47 7 + 4)^111 is expanded by the binomial formula all of its terms will be a multiple of 7 except the last which is 4^111.
4^111 = 64^37 = (9*7 + 1)^37 All of the terms when this is expanded are a multiple of 7 except the last which is 1. This shows that 333^111 = 7p + 1 where p is some integer.
Thus 111^333 + 333^111 = 7n - 1 + 7p + 1 = 7(n + p) where (n + p) is some integer so the product is exactly divisible by 7.