Find the digit in the units placr of 15^28+11^22-9^27

Consider 15²⁸.

Since 15 ends in 5, then so are 15ⁿ, ∀n ∈ N ∧ n ≠ 0.

Hence 15²⁸ also ends in 5.

Consider 11²².

Since 11 ends in 1, then so are 11ⁿ, ∀n ∈ N ∧ n ≠ 0.

Hence 11²² also ends in 1.

Consider 9²⁷.

As 9 is here, 9ⁿ end in 9 if n is odd and 9ⁿ end in 1 if n is even.

Since the exponent of 9 here, 27, is odd, 9²⁷ ends in 9.

So,

→  ones digit of 15²⁸ is 5.

→  ones digit of 11²² is 1.

→  ones digit of 9²⁷ is 9.

Thus,

15²⁸ + 11²² - 9²⁷  ≡  5 + 1 - 9  =  - 3  ≡  7 (mod 10)

Hence 7 is the answer.