The document contains code to calculate the sum of the digits of the factorial of a given number. It initializes an array to store the digits, calculates the factorial iteratively by multiplying by numbers from 1 to the given number, stores the digits in the array, prints out the factorial and calculates the sum of its digits.

1. #include<stdio.h>
void main()
{
int A[200]={0,1};
int i,j,k,h,n=2,s=0,m;
clrscr();
printf("Put a number for sum of the digits of its factorial.n");
scanf("%d",&m);
for(i=2;i<=m;i++)
{
h=0;
for(j=n-1;j>=0;j--)
{
k=A[j]*i+h;
A[j+2]=k%10;
h=k/10;
}
A[1]=h%10;
A[0]=h/10;
n=n+2;
}
printf("nnn");
for(i=0;i<2*m;i++)
s=s+A[i];
printf("The sum of the digits is %d",s);
getch();
}
SUM OF FACTORIAL DIGITS
1

2. #include<stdio.h>
void main()
{
int A[400]={1,0,2,4};
int i,j,k,h,n=4,s=0,m=100;
clrscr();
for(i=2;i<=m;i++)
{
h=0;
for(j=n-1;j>=0;j--)
{
k=A[j]*1024+h;
A[j+4]=k%10;
h=k/10;
}
A[3]=h%10;
h=h/10;
A[2]=h%10;
h=h/10;
A[1]=h%10;
A[0]=h/10;
n=n+4;
}
printf("nnn");
for(i=0;i<4*m;i++)
s=s+A[i];
printf("The sum of the digits is %d",s);
getch();
}
POWER DIGIT SUM
2

3. #include<stdio.h>
void main()
{
int i=0,j,p,temp,A[10]={0,1,2,3,4,5,6,7,8,9};
long int f,k=1,M=1000000;
long int fact(int);
clrscr();
while((k!=0)&&(i<9))
{
f=fact(9-i);
p=M/f;
k=M%f;
M=k;
if(p>0)
{
if(k==0)
p=p-1;
temp=A[p+i];
for(j=p+i;j>i;j--)
A[j]=A[j-1];
A[i]=temp;
}
i++;
}
printf("nnThe required number is:t");
for(j=0;j<i;j++)
printf("%d",A[j]);
for(j=9;j>=i;j--)
printf("%d",A[j]);
getch();
}
long int fact(int m)
{
long int r=1;
if(m==0)
return(r);
else
return(m*fact(m-1));
}
LEXICOGRAPHIC PERMUTATIONS
3

4. #include<stdio.h>
void main()
{
int n,m,i;
long int s=0;
int d(int);
clrscr();
printf("Input an upper bound.n");
scanf("%d",&n);
for(i=12;i<=n;i++)
{
m=d(i);
if((m>i)&&(i==d(m)))
{
if(m>n)
m=0;
s=s+m+i;
}
}
printf("nnThe sum is :%ld.",s);
getch();
}
int d(int a)
{
int r=0,j;
for(j=1;j<a;j++)
{
if(a%j==0)
r=r+j;
}
return(r);
}
AMICABLE NUMBERS
4

5. #include<stdio.h>
void main()
{
int A[200]={0,1};
int i,j,k,h,n=2,s=0,m;
clrscr();
printf("Put a number for sum of the digits of its factorial.n");
scanf("%d",&m);
for(i=2;i<=m;i++)
{
h=0;
for(j=n-1;j>=0;j--)
{
k=A[j]*i+h;
A[j+2]=k%10;
h=k/10;
}
A[1]=h%10;
A[0]=h/10;
n=n+2;
}
printf("nnn");
i=0;
while(1)
{
if(A[i]!=0)
break;
i++;
}
printf("nnThe factorial of %d is :nn",m);
for(j=i;j<2*m;j++)
{
printf("%d",A[j]);
s=s+A[j];
}
printf("nnThe number of digits is %d.",2*m-i);
printf("nnThe sum of the digits is %d",s);
getch();
}
SUM OF FACTORIAL DIGITS
5

6. Put a number for sum of the digits of its factorial.
100
The factorial of 100 is :
9332621544394415268169923885626670049071596826438162146859296389521759999322991
5608941463976156518286253697920827223758251185210916864000000000000000000000000
The number of digits is 158.
The sum of the digits is 648
Put a number for sum of the digits of its factorial.
20
The factorial of 20 is :
2432902008176640000
The number of digits is 19.
The sum of the digits is 54
Put a number for sum of the digits of its factorial.
10
The factorial of 10 is :
3628800
The number of digits is 7.
The sum of the digits is 27
OUTPUT-1
OUTPUT-2
OUTPUT-3
6

7. #include<stdio.h>
#include<math.h>
void main()
{
int A[404]={0},i,j,k,h,n,p,q,s=0,m,d;
clrscr();
printf("Input the power.n");
scanf("%d",&p);
q=p%10;
m=p/10;
d=pow(2,q);
if(q<4)
{
n=1;
A[0]=d;
}
else if(q<7)
{
n=2;
A[1]=d%10;
A[0]=d/10;
}
else
{
n=3;
A[2]=d%10;
d=d/10;
A[1]=d%10;
A[0]=d/10;
}
for(i=1;i<=m;i++)
{
h=0;
for(j=n-1;j>=0;j--)
{
k=A[j]*1024+h;
A[j+4]=k%10;
h=k/10;
}
A[3]=h%10;
h=h/10;
A[2]=h%10;
h=h/10;
A[1]=h%10;
A[0]=h/10;
n=n+4;
}
j=403;
while(1)
{
if(A[j]!=0)
break;
j--;
}
k=0;
while(1)
{
if(A[k]!=0)
break;
k++;
}
printf("nnnThe number 2 raised to %d is :nn",p);
for(i=k;i<=j;i++)
POWER DIGIT SUM
7

8. {
printf("%d",A[i]);
s=s+A[i];
}
printf("nnThe number of digits is %d.nn",j-k+1);
printf("The sum of the digits is %d",s);
getch();
}
8

9. Input the power.
1000
The number 2 raised to 1000 is :
107150860718626732094842504906000181056140481170553360744375038837035105112493612249
319837881569585812759467291755314682518714528569231404359845775746985748039345677748
242309854210746050623711418779541821530464749835819412673987675591655439460770629145
71196477686542167660429831652624386837205668069376
The number of digits is 302.
The sum of the digits is 1366
Input the power.
20
The number 2 raised to 20 is :
1048576
The number of digits is 7.
The sum of the digits is 31
Input the power.
5
The number 2 raised to 5 is :
32
The number of digits is 2.
The sum of the digits is 5
OUTPUT-1
OUTPUT-2
OUTPUT-3
9

10. #include<stdio.h>
void main()
{
int i=0,j,p,temp,A[10]={0,1,2,3,4,5,6,7,8,9};
long int f,k=1,M,q;
long int fact(int);
clrscr();
printf("nnInput the lexicographic ordinal status. ");
scanf("%ld",&M);
printf("nnInput the number of digits using. ");
scanf("%d",&q);
f=fact(q);
if(M>f)
printf("nnThere are only %ld numbers in the list.",f);
else
{
while((k!=0)&&(i<q-1))
{
f=fact(q-i-1);
p=M/f;
k=M%f;
M=k;
if(p>0)
{
if(k==0)
p=p-1;
temp=A[p+i];
for(j=p+i;j>i;j--)
A[j]=A[j-1];
A[i]=temp;
}
i++;
}
printf("nnThe required number is:t");
for(j=0;j<i;j++)
printf("%d",A[j]);
for(j=q-1;j>=i;j--)
printf("%d",A[j]);
}
getch();
}
long int fact(int m)
{
long int r=1;
if(m==0)
return(r);
else
return(m*fact(m-1));
}
LEXICOGRAPHIC PERMUTATIONS
10

11. Input the lexicographic ordinal status.
1000000
Input the number of digits using.
10
The required number is: 2783915460
Input the lexicographic ordinal status.
5
Input the number of digits using.
3
The required number is: 201
Input the lexicographic ordinal status.
4000000
Input the number of digits using.
10
There are only 3628800 numbers in the list.
OUTPUT-1
OUTPUT-2
OUTPUT-3
11

12. #include<stdio.h>
void main()
{
int n,m,i;
long int s=0;
int d(int);
clrscr();
printf("Input an upper bound.n");
scanf("%d",&n);
printf("nnThe amicable pairs are :nn");
for(i=12;i<=n;i++)
{
m=d(i);
if((m>i)&&(i==d(m)))
{
printf("%dt%dnn",i,m);
if(m>n)
m=0;
s=s+m+i;
}
}
printf("nnThe sum of the amicable numbers below %d is :%ld.",n,s);
getch();
}
int d(int a)
{
int r=0,j;
for(j=1;j<a;j++)
{
if(a%j==0)
r=r+j;
}
return(r);
}
AMICABLE NUMBERS
12

13. Input an upper bound.
10000
The amicable pairs are :
220 284
1184 1210
2620 2924
5020 5564
6232 6368
The sum of the amicable numbers below 10000 is :31626.
Input an upper bound.
250
The amicable pairs are :
220 284
The sum of the amicable numbers below 250 is :220.
Input an upper bound.
500
The amicable pairs are :
220 284
The sum of the amicable numbers below 500 is :504.
OUTPUT-1
OUTPUT-2
OUTPUT-3
13

14. #include<stdio.h>
#include<math.h>
int m=1;
void main()
{
long int p,q;
int i,j,k,n,r=1,A[5000]={1},s=0;
long int pro(int);
clrscr();
printf("Input a number for the sum of its factorial digits.nn");
scanf("%d",&n);
while(m<=n)
{
p=pro(n);
q=0;
for(k=0;k<r;k++)
{
q=A[k]*p+q;
A[k]=q%10;
q=q/10;
}
while(q!=0)
{
A[k]=q%10;
q=q/10;
k++;
}
r=k;
}
printf("nnThe factorial of %d is :nn",n);
for(i=r-1;i>=0;i--)
{
printf("%d",A[i]);
s=s+A[i];
}
printf("nnThe number of digits is %d.nn",r);
printf("The sum of the digits is %d.",s);
getch();
}
long int pro(int a)
{
long int l=1,b;
b=pow(2,26);
while((l<=b/m)&&(m<=a))
{
l=l*m;
m++;
}
return(l);
}
SUM OF FACTORIAL DIGITS
14

15. #include<stdio.h>
#include<math.h>
void main()
{
long int p,q;
int i,j=0,k,l,m,n=1,A[310]={1},s=0;
clrscr();
printf("Input the power.n");
scanf("%d",&m);
l=m/27;
p=pow(2,27);
for(i=0;i<2;i++)
{
while(j<l)
{
q=0;
for(k=0;k<n;k++)
{
q=A[k]*p+q;
A[k]=q%10;
q=q/10;
}
while(q!=0)
{
A[k]=q%10;
q=q/10;
k++;
}
n=k;
j++;
}
j--;
p=pow(2,m%27);
}
printf("The number 2 raised to %d is :nn",m);
for(i=n-1;i>=0;i--)
{
printf("%d",A[i]);
s=s+A[i];
}
printf("nnThe number of digits is %d.nn",n);
printf("The sum of the digits is %d.",s);
getch();
}
POWER DIGIT SUM
15