Question

write a program to check whether to accept a number and check whether the number is excessive or deficient or perfect number​

Answer 1

Answer:

STEP 1: START.

STEP 2: SET sum= 0.

STEP 3: REPEAT STEP 4 UNTIL i<=? n.

STEP 4: if (n% i==0) then. if(n/i==i) sum=sum+i. else. sum=sum+i. sum=sum+ (n/i)

STEP 5: RETURN sum.

STEP 6: END.

Explanation:

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Answer 2

Answer:

#include <bits/stdc++.h>

using namespace std;

int getSum(int n)

{

   int sum = 0;

   for (int i=1; i<=sqrt(n); i++)

   {

       if (n%i==0)

       {

           if (n/i == i)

               sum = sum + i;

           else  

           {

               sum = sum + i;

               sum = sum + (n / i);

           }

       }

   }

   sum = sum - n;

   return sum;

}

bool checkDeficient(int n)

{

   return (getSum(n) < n);

}

int main()

{

int n;

cout << "\n\n Check whether a given number is an Deficient number:\n";

cout << " --------------------------------------------------------\n";

cout << " Input an integer number: ";

cin >> n;

   checkDeficient(n)? cout << " The number is Deficient.\n" : cout << " The number is not Deficient.\n";

   return 0;

}

Explanation:

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