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Betrothed numbers in C Program?
Here we will see the Betrothed number. This is a pair of numbers, such that the sum of the proper divisors of one number is one more than the other number. We have to find these pairs
For an example, the pair is like (48, 75). So the divisors of 48 is {1, 2, 3, 4, 6, 8, 12, 16, 24} and sum is 76. Similarly, the divisors of 75 is {1, 3, 5, 15, 25} so sum is 49.
Algorithm
BetrothedPairs (n) −
begin for num in range 1 to n, do sum := 1 for i in range 2 to num, do if num is divisible by i, then sum := sum + i if i * i is not same as num, then sum := sum + num / i end if end if if sum > num, then num2 := sum – 1 sum2 := 1 for j in range 2 to num2, do if num2 is divisible by j, then sum2 := sum2 + j if j * j is not same as num2, then sum2 := sum2 + num2 / j end if end if done if sum2 = num + 1, then print the pair num and num2 end if end if done done end
Example
#include <iostream>
using namespace std;
void BetrothedPairs(int n) {
for (int num = 1; num < n; num++) {
int sum = 1;
for (int i = 2; i * i <= num; i++) { //go through each number to get proper divisor
if (num % i == 0) {
sum += i;
if (i * i != num) //avoid to include same divisor twice
sum += num / i;
}
}
if (sum > num) {
int num2 = sum - 1;
int sum2 = 1;
for (int j = 2; j * j <= num2; j++){
if (num2 % j == 0) {
sum2 += j;
if (j * j != num2)
sum2 += num2 / j;
}
}
if (sum2 == num+1)
cout << "(" << num << ", " << num2 <<")" << endl;
}
}
}
int main() {
int n = 5000;
BetrothedPairs(n);
}Output
1
Updated on: 20-Aug-2019
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