Given a number N. The task is to find the count of numbers which have N digits and even number of zeroes.
Note: The number can have preceding 0’s.
Input: N = 2 Output: Count = 81 Total 2 digit numbers are 99 considering 1 as 01. 2 digit numbers are 01, 02, 03, 04, 05.... 99 Numbers with odd 0's are 01, 02, 03, 04, 05, 06, 07, 08, 09 10, 20, 30, 40, 50, 70, 80, 90 i.e. 18 The rest of the numbers between 01 and 99 will do not have any zeroes and zero is also an even number. So, numbers with even 0's are 99 - 18 = 81. Input: N = 3 Output: Count = 755
Approach: The idea is to find the Count Numbers with N digits which consists of odd number of 0’s and subtract it from the total number with N digits to get the number with even 0’s.
Note: Answer can be very large, so for N greater than 9, use modular exponentiation.
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