perfect,abundant,non-abundant and prime numbers

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 :\n");

for (int i = 2; i < 10000; i++)

{

factors.Clear();

factors.Add(1);

for(int j = 2; j <= i / 2 ; j++)

{

if (i % j == 0) factors.Add(j);

}

int sum = 0;

for (int k = 0; k < factors.Count; k++)

{

sum += factors[k];

}

if (sum == i)

{

for (int l = 0; l < factors.Count - 1; l++)

{

Console.Write("{0} + ", factors[l]);

}

Console.Write("{0} = {1}\n", 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 :\n");

for (int i = 2; i < 100; i++)

{

factors.Clear();

factors.Add(1);

for(int j = 2; j <= i / 2 ; j++)

{

if (i % j == 0) factors.Add(j);

}

int sum = 0;

for (int k = 0; k < factors.Count; k++)

{

sum += factors[k];

}

if (sum > i)

{

for (int l = 0; l < factors.Count - 1; l++)

{

Console.Write("{0} + ", factors[l]);

}

Console.Write("{0} > {1}\n", 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 :\n");

for (int i = 1; i < 100; i++) // start from 1

{

factors.Clear();

factors.Add(1);

for(int j = 2; j <= i / 2 ; j++)

{

if (i % j == 0) factors.Add(j);

}

int sum = 0;

for (int k = 0; k < factors.Count; k++)

{

sum += factors[k];

}

if (sum <= i)

{

for (int l = 0; l < factors.Count - 1; l++)

{

Console.Write("{0} + ", factors[l]);

}

Console.Write("{0} <= {1}\n", factors[factors.Count - 1], i);

}

using System;

using System.Collections.Generic;

class Prime

{

static void Main()

{

Console.WriteLine("The prime numbers less than 100 are :\n");

for (int i = 2; i < 100; i++)

{

bool isPrime = true;

for(int j = 2; j <= i / 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.

4 Replies

Know the answer? Post it — somebody with the same question will find it here.

Vulpes15y ago
Accepted 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 Mxoli15y ago
Thanks alot!!!
Avuya Mxoli15y ago
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

}
Vulpes15y ago
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 :\n");

for (int i = 2; i < 10000; i++)
{
factors.Clear();
factors.Add(1);
for(int j = 2; j <= i / 2 ; j++)
{
if (i % j == 0) factors.Add(j);
}
int sum = 0;
for (int k = 0; k < factors.Count; k++)
{
sum += factors[k];
}
if (sum == i)
{
for (int l = 0; l < factors.Count - 1; l++)
{
Console.Write("{0} + ", factors[l]);
}
Console.Write("{0} = {1}\n", 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 :\n");

for (int i = 2; i < 100; i++)
{
factors.Clear();
factors.Add(1);
for(int j = 2; j <= i / 2 ; j++)
{
if (i % j == 0) factors.Add(j);
}
int sum = 0;
for (int k = 0; k < factors.Count; k++)
{
sum += factors[k];
}
if (sum > i)
{
for (int l = 0; l < factors.Count - 1; l++)
{
Console.Write("{0} + ", factors[l]);
}
Console.Write("{0} > {1}\n", 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 :\n");

for (int i = 1; i < 100; i++) // start from 1
{
factors.Clear();
factors.Add(1);
for(int j = 2; j <= i / 2 ; j++)
{
if (i % j == 0) factors.Add(j);
}
int sum = 0;
for (int k = 0; k < factors.Count; k++)
{
sum += factors[k];
}
if (sum <= i)
{
for (int l = 0; l < factors.Count - 1; l++)
{
Console.Write("{0} + ", factors[l]);
}
Console.Write("{0} <= {1}\n", factors[factors.Count - 1], i);
}
}

}
}

using System;
using System.Collections.Generic;

class Prime
{
static void Main()
{
Console.WriteLine("The prime numbers less than 100 are :\n");

for (int i = 2; i < 100; i++)
{
bool isPrime = true;

for(int j = 2; j <= i / 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.