**Concept: The question tests on the basics of** **Unit Digit**

**Solution:**

The cyclicity of 1 is 1 and the cyclicity of 9 is 2

Therefore 171^125 will have its units digit as 1

For 29^125, we divide the power 125 by the cyclicty which is 2, to get a remainder of 1.

So the last digit is 9^1=9

When 1 is subtracted from 0, we carry over 1 from the 10's place. We get 11 - 9 = 2**Option B**