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