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C++ program to find nearest integer for which the number and its digits sum gcd is greater than 1
Suppose we have a number N. Consider a function gcdSum(x) of a positive integer x which is the gcd of that integer with its sum of digits. We have to find the smallest integer x >= n, such that gcdSum(x) > 1.
So, if the input is like N = 31, then the output will be 33, because gcd of 31 and (3+1) is 1. The gcd of 32 and (3+2) is 1, and gcd of 33 and (3+3) is 3.
Steps
To solve this, we will follow these steps −
for initialize i := n, when i <= n + 2, update (increase i by 1), do: jml := 0 x := i while x > 0, do: jml := jml + x mod 10 x := x / 10 if gcd of i and jml is not equal to 1, then: return i return 0
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
int solve(int n) {
for (long i = n; i <= n + 2; i++) {
long jml = 0;
long x = i;
while (x > 0) {
jml += x % 10;
x /= 10;
}
if (__gcd(i, jml) != 1) {
return i;
}
}
return 0;
}
int main() {
int N = 31;
cout << solve(N) << endl;
}Input
31
Output
33
Updated on: 03-Mar-2022
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