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 one year ago
Write a program which given an integer n, classifies it as 'perfect', 'abundant' or 'deficient'. A number is perfect if the sum of its divisors ( excluding the number itself) is equal to the number, e.g. 6. A number is abundant if the sum of its divisors(excluding the number itself) is greater than the number, e.g. 12. A number is deficient if it is neither perfect nor abundant
 one year ago
Write a program which given an integer n, classifies it as 'perfect', 'abundant' or 'deficient'. A number is perfect if the sum of its divisors ( excluding the number itself) is equal to the number, e.g. 6. A number is abundant if the sum of its divisors(excluding the number itself) is greater than the number, e.g. 12. A number is deficient if it is neither perfect nor abundant

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time_to_learn
 one year ago
Best ResponseYou've already chosen the best response.0What do you need help with in this problem?

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0Give at least a pseudocode so that I can understand what to do

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0you should try this program out yourself first, before asking people to do it for you.

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0#include <iostream> 08 using namespace std; 09 10 int main() 11 { 12 int number,counter,total; 13 14 /* Asks for user input*/ 15 cout << "Welcome to the program Factors, this program check to see\n" 16 << "if the number you enter is deficent, perfect or abundant\n" 17 << "this program will read in numbers till you enter 1 and or negitive numbers"; 18 19 while (number > 1) 20 { 21 22 counter = 1; 23 cout << endl; 24 cout << "Please enter a Integer:\n"; 25 cin >> number; 26 cout << endl; 27 28 29 30 if (number % counter == 0) 31 { 32 total= counter + (number % counter); 33 34 cout << total << "+"; 35 counter++; 36 } 37 38 39 40 if (number < total) 41 { 42 cout << endl; 43 cout << number <<" is abundant\n"; 44 45 } 46 else if (number>total) 47 { 48 cout << endl; 49 cout << number << " is deficient\n"; 50 51 cout << endl; 52 } 53 else 54 { 55 cout << endl; 56 cout << number << " is Perfect\n"; 57 58 cout << endl; 59 } 60 61 } 62 63 cout<< endl << endl << endl; 64 system ("pause"); 65 return 0; 66 }

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0I mean, what are they?

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0I think something is wrong with my values?

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0But i cannot put my finger on it...

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0The thing which is annoying is you have to add up EVERY possible combination. So for 12 it's 3 + 4 + 2 + 6....

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0its like adding prime factors of 12 excluding itself?

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0so if it is working on a number that isnt perfect It will work through the IF statements to see if it is abundant and another to see if it is deficient

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0I will post the code properly

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0#include <iostream> using namespace std; int main() { int number = 0,counter,total; /* Asks for user input*/ cout << "Welcome to the program Factors, this program check to see\n" << "if the number you enter is deficent, perfect or abundant\n" << "this program will read in numbers till you enter 1 and or negitive numbers"; while (number > 1) { cout << endl; cout << "Please enter a Integer:\n"; cin >> number; if (number < 0) { break; } cout << endl; counter = 1; total = 0; while (counter < number) { if (number % counter == 0) { cout << total << "+" << counter; total += counter; cout << "=" << total << endl; } counter++; } if (number < total){ cout << endl; cout << number <<" is abundant\n"; } else if (number>total) { cout << endl; cout << number << " is deficient\n"; cout << endl; } else { cout << endl; cout << number << " is Perfect\n"; cout << endl; } } cout<< endl << endl << endl; system ("pause"); return 0; }

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0#include <iostream> using namespace std; /* */ int main () { int highNum, abunTotal, defTotal, perfTotal; abunTotal = 0; // sum of divisors more than the number itself defTotal = 0; // sum of divisors less than the number itself perfTotal = 0; // sum of divisors equal to the number itself int sum = 0; cout<< "Please enter a positive integer:"; cin>> highNum; for(int i = 1; i <= highNum; i++ ) { sum = 0; for(int j = 1; j < i; j++ ) { if((i % j) == 0) { sum += j; } } if ( sum == i) //the number is perfect { perfTotal++; cout << "PERFECT NUMBER: " << i << endl; } if ( sum < i) //the number is deficient { defTotal++; } if ( sum > i) //the number is abundant { abunTotal++; } } cout << "The perfTotal is:" << perfTotal << endl; cout << "The defTotal is:" << defTotal << endl; cout << "The abunTotal is:"<< abunTotal << endl; cout << endl; }
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