Question

Best answer will be marked brainliest

Answer 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.

It's helps you a lot .