//program to print/display ...11111,1111,111,11,1 to nth term.

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

int m=1,n,number,denom;

printf("enter value of 'n' as a number of terms \n");

scanf("%d",&n);

printf("now enter number having %d digits(1)\n",n)

scanf("%d",&number);while(n>=m)

{

printf("%d\n",number);

denom=pow(10,n-1);

number=number/denom;

n--;

}

getch();

}

if n= 6(It means a number with 6 ones i.e. 111111)

111111,11111,1111,111,...

can be shown as

first 11111

second dividing above number by 10th power of entered number -1

as the second iteration goes , it has value with 1 less digit.

it continues until the value lasts to 1.

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

int m=1,n,number,denom;

printf("enter value of 'n' as a number of terms \n");

scanf("%d",&n);

printf("now enter number having %d digits(1)\n",n)

scanf("%d",&number);while(n>=m)

{

printf("%d\n",number);

denom=pow(10,n-1);

number=number/denom;

n--;

}

getch();

}

**..................................****logics in mind:**if n= 6(It means a number with 6 ones i.e. 111111)

111111,11111,1111,111,...

can be shown as

first 11111

second dividing above number by 10th power of entered number -1

as the second iteration goes , it has value with 1 less digit.

it continues until the value lasts to 1.

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