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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
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