Shivam, I hope I am not too late to help. Let's look at what we have.
Z has to be odd, because the last digit of our product is 5, so Z can be 1, 3, 5, 7 or 9.
1 won't work, because XY5 x 1 = XY5.
Let's try Z = 3. Then 3 x 5 = 15. We have our 5 in the product and we carry over a 1. 3 x Y + now has to end in a 1. There is no single digit that we can multiply 3 by, add one, and end in 1, so 3 is out.
Z = 5. 5 x 5 = 25. There's our 5 and carry the 2. 5 x Y +2 now has to end in a 1. Not possible, since anything will end in a 2 or 7.
Z = 7. 7 x 5 = 35. There's our 5 again, carry the 3. Now, 7 x Y +3 has to end in a 1. Hey, Y could be 4! 7 x 4 + 3 = 31. There's our 1 and carry the 3. Now, 7 x X + 3 has to be X2. The only way we can end in a 2 is to make X = 7. 7 x 7 + 3 = 52, but that doesn't quite work.
We are left with Z = 9. 9 x 5 = 45. There's our 5, carry the 4. We need 9 x Y + 4 to end in a 1. 9 x 3 + 4 = 31, so Y is 3. There's our 1, carry the 3. 9 x X + 3 has to end in a 2 and start with X. Wouldn't ya know it, 9 x 1 + 3 = 12, so X = 1.
So, we have 135 x 9 = 1215. Your answer is 9!
