Given two numbers m and n, count number of m digit numbers that are divisible by n.
Input : m = 2 n = 6 Output : 15 Two digit numbers that are divisible by 6 are 12, 18, 24, 30, 36, ....., 96. Input : m = 3 n = 5 Output :180
A simple solution is two try all m digit numbers. For every number, check if it is divisible by n. If yes, we increment count.
An efficient solution involves following steps.
The idea is based on the fact that starting from first divisible number, every n-th number is divisible by n.
- Find largest m digit number.
- Find largest m-1 digit number.
- Divide both number by n and subtract later from prior.
Below is the implementation of above steps.
Time complexity : O(m)
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