n and n+1 are integers with the same number of positive divisions. Find the integers n from 1
n and n+1 are integers with the same number of positive divisions. Find the integers n from 1<n<107. For example, the positive divisors of 14 are 1, 2, 7, 14, and 15 are 1, 3, 5, 15.
(P.s.: You have to done it by C++.)
Programming instructions:
- Include the necessary header file.
- In the main function, declare the required variables.
- Use for-loop to call the divisorcount() function for each number.
- Call the divisorcount() function for a number and store the value returned by the function in a variable.
- Again all the divisorcount() function for a number less than the previous one store the value returned by the function in a variable.
- Compare both the variables.
- If they are equal print them.
- The divisorcount() function counts the number of divisors for a number.
Program:
// Header file
#include<stdio.h>
int divisorcount(int);
//Main function
int main()
{
//Declare variables
int n,divisor1,divisor2,j=0;
//for-loop to call the function several times
for(n=1;n<=107;n++){
//Call the function for a number to count the number of divisors
divisor1=divisorcount(n);
//Call the same function for a number less than previous one
divisor2=divisorcount(n-1);
//Compare the numbers
if(divisor1==divisor2){
printf("%d and %d\n",n-1,n);
j++;
}
}
printf("\nThere are %d integers.",j);
}
//Function to count the number of divisors
int divisorcount(int num)
{
int i,divi=0;
for(i=1;i<=(num)/2;i++)
if(num%i==0)
divi++;
return divi;
}
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