Counting by 2’s


Write a C++ program to count the number of 2's between 0 and (n) where (n) is a positive integer.


We need two loops, one goes from 0 to n and the other one examines the number digit by digit. One way to get the decimal digits of a number is to shift the number to the right. Shifting is nothing but a division by 10. The current digit can be extracted by taking the remainder of the division operation. Here is an example:

x = 123

d = x%10 = 3
x = x/10 = 12

d = 12%10 = 2
x = x/10 = 1

d = x%10 = 1
x = x/10 = 0

This is not an efficient algorithm. In the next article we will solve it in a better way.


Here is the code in C++

#include <iostream>

using namespace std;

//Returns number of
//2's from 0 to n
int NumTwos(int n)
	//Total count of 2's
	int count = 0;
	//For each number in range
	for (int i = 0; i <= n; i++)
		//Work on a copy of the 
		//numbe because we are 
		//going to change it
		int x = i;
		//Keep dividing the number 
		//by 10 until you get 0
		while (x != 0)
			//The first digit from the 
			//right is the remainder of 
			//the division
			int d = x%10;

			//Shift to right by dividing by 10
			x = x/10;
			//If the digit is 2 then 
			//increment the 2's counter
			if (d == 2)

	return count;

//Main function
void main()
	cout << NumTwos(1);
	cout << NumTwos(10);
	cout << NumTwos(20);
	cout << NumTwos(30);
	cout << NumTwos(100);

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