perfectabundantnon-abundant and prime numbers

Question

Hello Plz help me wth pseudocode of the above mentioned and also determing whether a number is a divisor of another number. If you are not sure abt the pseudocode u can give me c# code il try and convert it. thnks

Vulpes · Answer

Sure for the prime number I'd suggest: FUNCTION IsPrime(integer:n) boolean prime = true; FOR j goes from 2 to (n/2) If n MOD j EQUALS 0 prime = false; EXIT FOR; END IF END FOR IF prime EQUALS true WRITE(prime number); ELSE WRITE(not a prime number); END IF END FUNCTION

Avuya Mxoli · Answer

Thanks alot!!!

Avuya Mxoli · Answer

thanks alot...I'll translate that to pseudocode but cud u plz just show me if maybe the prime number one were to be a function... I have something like for perfect number: function IsPerfect(integer:n) { integer sum =0; integer j; FOR j goes from 1 to (n/2) IF n MOD j EQUALS 0 sum = sum + j; END IF END FOR IF sum EQUALS n WRITE(perfect number); ELSE WRITE(not a perfect number) END FUNCTION }

Vulpes · Answer

As I'm a bit short on time just now and I'm not too good at pseudo-code anyway I'll give you the C# code for each of those. The first three are very similar (as I've never heard of it before I assume a non-abundant number is a number which is not abundant): using System; using System.Collections.Generic; class Perfect { static void Main() { List factors = new List (); Console.WriteLine(The perfect numbers less than 10000 are :
); for (int i = 2; i { factors.Clear(); factors.Add(1); for(int j = 2; j { if (i % j == 0) factors.Add(j); } int sum = 0; for (int k = 0; k { sum += factors[k]; } if ( sum == i ) { for (int l = 0; l { Console.Write({0} +  factors[l]); } Console.Write({0} = {1}
 factors[factors.Count - 1] i); } } } } using System; using System.Collections.Generic; class Abundant { static void Main() { List factors = new List (); Console.WriteLine(The abundant numbers less than 100 are :
); for (int i = 2; i { factors.Clear(); factors.Add(1); for(int j = 2; j { if (i % j == 0) factors.Add(j); } int sum = 0; for (int k = 0; k { sum += factors[k]; } if (sum > i ) { for (int l = 0; l { Console.Write({0} +  factors[l]); } Console.Write({0} > {1}
 factors[factors.Count - 1] i); } } } } using System; using System.Collections.Generic; class NonAbundant { static void Main() { List factors = new List (); Console.WriteLine(The non-abundant numbers less than 100 are :
); for (int i = 1; i start from 1 { factors.Clear(); factors.Add(1); for(int j = 2; j { if (i % j == 0) factors.Add(j); } int sum = 0; for (int k = 0; k { sum += factors[k]; } if ( sum ) { for (int l = 0; l { Console.Write({0} +  factors[l]); } Console.Write({0} } } } } using System; using System.Collections.Generic; class Prime { static void Main() { Console.WriteLine(The prime numbers less than 100 are :
); for (int i = 2; i { bool isPrime = true; for(int j = 2; j { if (i % j == 0) { isPrime = false; break; } } if(isPrime) Console.WriteLine(i); } } } All of these use a straightforward 'brute force' approach rather than some of the 'clever' methods which exist for generating these numbers more quickly.

perfect,abundant,non-abundant and prime numbers