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# Count Numbers with N digits which consists of odd number of 0's in C++

We are given a number N as input. The goal is to find all N digit numbers that have an odd number of 0’s as digits. The number also could have preceding zeros like in case of N=3 numbers included will be 000,011,012….990.

Let us understand with examples.

**Input** − N=3

**Output** − Count of no. with N digits which consists of even number of 0's are − 244

**Explanation** − All 3 digit numbers would be like −

Smallest will be 000, then 011,012,013,0014…..Highest will be 990.

**Input** − N=5

**Output** − Count of no. with N digits which consists of even number of 0's are − 33616

**Explanation** − All 5 digit numbers would be like −

Smallest will be 00000, then 00011,00012,00013,0014…..Highest will be 99990.

## The approach used in the below program is as follows

We will first calculate the total N digit numbers that are T=10N-1. Then calculate all N digit numbers with even 0’s as digits, that is E=10N-8N . The remaining numbers with Odd0’s in digits will be (T-E)/2.

Take an integer N as input.

Function count_dd(int N) takes N and returns the count of N digit numbers with odd 0’s.

Total N digit numbers are total=pow(10,N)-1

Total N digit numbers with even 0’s in digit are even=pow(10,N)-pow(8,N).

Leftover odd 0’s in digit are odd=(total-even)/2.

Return odd as a count of N digit numbers with odd numbers of 0’s.

## Example

#include <bits/stdc++.h> using namespace std; int count_odd(int N){ int total = pow(10, N); int even = pow(8, N); int odd = (total - even) / 2; return odd; } int main(){ int N = 4; cout<<"Count of Numbers with N digits which consists of odd number of 0's are: "<<count_odd(N); return 0; }

## Output

If we run the above code it will generate the following output −

Count of Numbers with N digits which consists of odd number of 0's are: 2952