Factorial of numbers

cior · #1 08-Jun-2004, 18:02

I've written a program to find all numbers where the factorial of the individual digits is the same as the number itself.
Example.

145 - 1!+4!+5! = 1+24+120 = 145

The program looks like this but it only seems to find numbers 1,2 & 145 but I'm sure there are more.

CPP / C++ / C Code:

#include <iostream>
#include <conio>
#include <math>
#include <stdlib>

int check(int number);
int main()
{
int i=1,summ=0;

while(i<9999999)
{

summ=check(i);
if(summ==i)
	cout << " " << i << " ";
i++;
}
cout << "DONE";

getch();
}


int check(int number)
{
int i=1,result=1,summ=0,digit1=0,digit2=0,digit3=0,digit4=0,digit5=0,digit6=0,digit7=0,digit8=0;
digit1=number/10000000;
digit2=number/1000000-(10*digit1);
digit3=number/100000-(100*digit1)-(10*digit2);
digit4=number/10000-(1000*digit1)-(100*digit2)-(10*digit3);
digit5=number/1000-(10000*digit1)-(1000*digit2)-(100*digit3)-(10*digit4);
digit6=number/100-(100000*digit1)-(10000*digit2)-(1000*digit3)-(100*digit4)-(10*digit5);
digit7=number/10-(1000000*digit1)-(100000*digit2)-(10000*digit3)-(1000*digit4)-(100*digit5)-(10*digit6);
digit8=number/1-(10000000*digit1)-(1000000*digit2)-(100000*digit3)-(10000*digit4)-(1000*digit5)-(100*digit6)-(10*digit7);

if(digit1!=0)
{
for (result=1, i=1; digit1>=i; i++)
	result=result*=i;
summ+=result;
}

if(digit2!=0)
{
for (result=1, i=1; digit2>=i; i++)
	result=result*=i;
summ+=result;
}

if(digit3!=0)
{
for (result=1, i=1; digit3>=i; i++)
	result=result*=i;
summ+=result;
}

if(digit4!=0)
{
for (result=1, i=1; digit4>=i; i++)
	result=result*=i;
summ+=result;
}

if(digit5!=0)
{
for (result=1, i=1; digit5>=i; i++)
	result=result*=i;
summ+=result;
}

if(digit6!=0)
{
for (result=1, i=1; digit6>=i; i++)
	result=result*=i;
summ+=result;
}

if(digit7!=0)
{
for (result=1, i=1; digit7>=i; i++)
	result=result*=i;
summ+=result;
}

if(digit8!=0)
{
for (result=1, i=1; digit8>=i; i++)
	result=result*=i;
summ+=result;
}

return summ;
}

I'm not really good at coding so I'm sure it's "ugly" but does anyone see anything wrong with the code.

WaltP · #2 08-Jun-2004, 21:04

Quote:

Originally Posted by cior

I've written a program to find all numbers where the factorial of the individual digits is the same as the number itself.
Example.

145 - 1!+4!+5! = 1+24+120 = 145

The program looks like this but it only seems to find numbers 1,2 & 145 but I'm sure there are more.

Why are you sure there are more? I'll write a version and see what I come up with.

Max Payne · #3 08-Jun-2004, 21:20

Quote:

Originally Posted by WaltP

Why are you sure there are more? I'll write a version and see what I come up with.

Hi,
I just found out that they are unique factorials and here is all of those that exists:

1 = 1!
2 = 2!
145 = 1! + 4! + 5!
40585 = 4! + 0! + 5! + 8! + 5!

there is no other integers like it (I think, I'm not a math guy) according to this:
unique Factorials

WaltP · #4 08-Jun-2004, 21:32

Quote:

Originally Posted by Max Payne

Hi,
I just found out that they are unique factorials and here is all of those that exists:

1 = 1!
2 = 2!
145 = 1! + 4! + 5!
40585 = 4! + 0! + 5! + 8! + 5!

there is no other integers like it (I think, I'm not a math guy) according to this:
unique Factorials

I just tested it for numbers up to 2000 and there were only the 3 mentioned.

I also just confirmed 40585 too.

I just tested thru 9999999 and those are the only 4.

Max Payne · #5 08-Jun-2004, 21:50

Well that means you don't need a program then...

anyway, cior, your program works well and it displays all the 4 numbers although it took some time to get the last number (40585). Maybe we should focus on efficiency to get the numbers real fast.

still, good work cior for able to code it..

cior · #6 09-Jun-2004, 08:15

Thanks, I found out why the code doesn't work for me, 0! is actually 1 not 0. But the problem is that the I had to skip the inital zeroes and because of that I accidentally also skipped the zeroes within the number.

cior · #7 09-Jun-2004, 08:36

Ok, I managed to rewrite the code to check if the next digit is higher than zero, if that's true 0! will be counted as 1 otherwise as 0.

CPP / C++ / C Code:

#include <iostream>
#include <conio>
#include <math>
#include <stdlib>

int check(int number);
int main()
{
int i=1,summ=0;

while(i<9999999)
{

summ=check(i);
if(summ==i)
	cout << " " << i << " ";
i++;
}
cout << "DONE";

getch();

}


int check(int number)
{
int i=1,result=1,summ=0,digit1=0,digit2=0,digit3=0,digit4=0,digit5=0,digit6=0,digit7=0,digit8=0;
digit1=number/10000000;
digit2=number/1000000-(10*digit1);
digit3=number/100000-(100*digit1)-(10*digit2);
digit4=number/10000-(1000*digit1)-(100*digit2)-(10*digit3);
digit5=number/1000-(10000*digit1)-(1000*digit2)-(100*digit3)-(10*digit4);
digit6=number/100-(100000*digit1)-(10000*digit2)-(1000*digit3)-(100*digit4)-(10*digit5);
digit7=number/10-(1000000*digit1)-(100000*digit2)-(10000*digit3)-(1000*digit4)-(100*digit5)-(10*digit6);
digit8=number/1-(10000000*digit1)-(1000000*digit2)-(100000*digit3)-(10000*digit4)-(1000*digit5)-(100*digit6)-(10*digit7);

//1
if(digit1==0)
{

}
else
{
for (result=1, i=1; digit1>=i; i++)
	result=result*=i;
summ+=result;
}
//2
if(digit2==0 && digit1!=0)
{
for (result=1, i=1; digit2>=i; i++)
	result=result*=i;
summ+=result;
}
else if(digit2==0)
{

}
else
{
for (result=1, i=1; digit2>=i; i++)
	result=result*=i;
summ+=result;
}
//3
if(digit3==0 && digit2!=0)
{
for (result=1, i=1; digit3>=i; i++)
	result=result*=i;
summ+=result;
}
else if(digit3==0)
{

}
else
{
for (result=1, i=1; digit3>=i; i++)
	result=result*=i;
summ+=result;
}
//4
if(digit4==0 && digit3!=0)
{
for (result=1, i=1; digit4>=i; i++)
	result=result*=i;
summ+=result;
}
else if(digit4==0)
{

}
else
{
for (result=1, i=1; digit4>=i; i++)
	result=result*=i;
summ+=result;
}
//5
if(digit5==0 && digit4!=0)
{
for (result=1, i=1; digit5>=i; i++)
	result=result*=i;
summ+=result;
}
else if(digit5==0)
{

}
else
{
for (result=1, i=1; digit5>=i; i++)
	result=result*=i;
summ+=result;
}
//6
if(digit6==0 && digit5!=0)
{
for (result=1, i=1; digit2>=i; i++)
	result=result*=i;
summ+=result;
}
else if(digit6==0)
{

}
else
{
for (result=1, i=1; digit6>=i; i++)
	result=result*=i;
summ+=result;
}
//7
if(digit7==0 && digit6!=0)
{
for (result=1, i=1; digit7>=i; i++)
	result=result*=i;
summ+=result;
}
else if(digit7==0)
{

}
else
{
for (result=1, i=1; digit7>=i; i++)
	result=result*=i;
summ+=result;
}
//8
if(digit8==0 && digit7!=0)
{
for (result=1, i=1; digit8>=i; i++)
	result=result*=i;
summ+=result;
}
else if(digit8==0)
{

}
else
{
for (result=1, i=1; digit8>=i; i++)
	result=result*=i;
summ+=result;
}


return summ;
}

WaltP · #8 09-Jun-2004, 20:08

Now that you got it working, here's my solution:

CPP / C++ / C Code:

#include <stdio.h>
#include <conio.h>

int main()
{
    int num;    // number being tested
    int nnum;   // temporary version of num
    int val;    // the total value of the factorials
    int fact;   // a factorial
    int n;      // loop counter to compute the factorial
    int nn;     //  don't remember....

    for (num = 1; num < 9999999; num++)
    {
        nnum = num;             // Save the value in the temp
        val  = 0;               // Initialize the total
        while (nnum > 0)        // While theres still some number left...
        {
            n = nnum % 10;      // Get the ones digit
            nn = n;
            nnum /= 10;         // Remove that digit
            fact = 1;           // Factorials start at 1  (0! = 1)
            while (n > 0)       // Loop until the digit is gone
            {
                fact = fact * n;    // Compute the factorial
                n--;                // next digit
            }
            val += fact;        // Add this factorial to the total
        }
        if (num == val)         // If the number and the total are the same...
                printf("%d \n", num);
    }
    
    return 0;
}

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