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Find Nth Positive Number Whose Digital Root is X in C++
In this problem, we are given two integer values N and X. Our task is to create a program to Find Nth positive number whose digital root is X.
Digital Root (X) is a single digit positive number which is found by adding digits of N recursively adding digits, till the sum becomes single digit.
Let’s take an example to understand the problem,
Input
N = 5, X = 4
Output
40
Solution Approach
A simple way to solve the problem is by counting the numbers with a digital root is X. For this, we will start from 1 and then check if the current number’s digital root is equal to X and count the numbers and return Nth such number.
Program to illustrate the working of our solution,
Example
#include <iostream> using namespace std; int calcDigitalRoot(int num) { int digitSum = 1000, number = num; while (digitSum >= 10) { digitSum = 0; while (number > 0) { digitSum += number % 10; number /= 10; } number = digitSum; } return digitSum; } int calcNthDigitalRoot(int X, int N) { int countDigitalRootVals = 0; for (int i = 1; countDigitalRootVals < N; ++i) { int digitalRoot = calcDigitalRoot(i); if (digitalRoot == X) { ++countDigitalRootVals; } if (countDigitalRootVals == N) return i; } return -1; } int main(){ int X = 4, N = 5; cout<<N<<"th positive number whose digital root is "<<X<<" is "<<calcNthDigitalRoot(X, N); return 0; }
Output
5th positive number whose digital root is 4 is 40
Efficient approach
An efficient approach to solve the problem is by finding the Nth number with digital root equal to X using the direct formula which is,
Nth number = (N + 1)*9 + X
Program to illustrate the working of our solution,
Example
#include <iostream> using namespace std; int calcNthDigitalRoot(int X, int N) { int nthNumber = (((N - 1) * 9) + X); return nthNumber; } int main() { int X = 4, N = 12; cout<<N<<"th positive number whose digital root is "<<X<<" is "<<calcNthDigitalRoot(X, N); return 0; }
Output
12th positive number whose digital root is 4 is 103