Given a positive integer X and Y, the task is to find the last digit of X in the given base Y.
Input: X = 10, Y = 7 Output: 3 10 is 13 in base 9 with last digit 3 Input: X = 55, Y = 3 Output: 1 55 is 3 in base 601 with last digit 1
- When we try to convert X into the base Y
- We repeatedly divide X by base Y and store the remainder.
- So the final result comprises of the remainders in the order of division steps.
- Lets say the remainder of division step 1 is p, step 2 is q, step 3 is r
- Then the resultant number in base Y will be rqp
- And the last digit will be p
- Therefore, we just need to find the first remainder of X when divided by Y to get the lsat digit in X in base Y.
last digit = X % Y
Below is the implementation of the above approach:
Time Complexity: O(1)
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