# What is strong number n Armstrong number in C language?

# Write a Program to find the fibonnacci series of n numbers in C language?

include.
include.
void fibo(int);.
void main().
{.
int num;.
clrscr();.
printf("\n\t Enter number of elements in series : ");.
scanf("%d",&num);.
if(num>0).
fibo(num…);.
else.
printf("\n\t Please enter positive number ");.
}.
void fibo(int num).
{.
int a=0,b=1,c=0;.
if(num==1).
printf("\n%d",a);.
if(num>=2).
printf("\n%d\t%d\t",a,b);.
for(;num>2;num--).
{.
c=a+b;.
a=b;.
b=c;.
printf("%3d\t",c);.
}.
getch();.

# Write a program in c language to check whether a given integer is a strong number or not using nested loops?

include void main() { int n,t,sum=0; float fact=1; clrscr(); printf("enter the number"); scanf(%d",&n); t=n; while(t) { for(i=0;i

# What is a Armstrong number?

702-934-4268

# What is a strong number?

As to your question I'm a bit puzzled. 2 is larger than 1 and 20 is more than 10 but less than 30. Alright, so you already know this much I'm assuming your question may relate… to the need for a strong password where there is a need for password protection. If this is the case, the need for a large combination of letters and numbers (alpha/numeric sequence) is the same as stated above. Use as large a seuence as may be permitted for login. Some people suggest you keep a book nearby the computer to help you construct passwords and this can be a useful technique. Try it out: Open the book and begin your password with the page number followed by the first letter of the each word in the first sentence (Use Capital letters if that is what you read and lower case letters if that is what you read. Or use the last sentence on the page or the first letter of each paragraph on the page. Again, the longer and larger is the combination of your alpha/numeric sequencing the Stonger is the Password.

# Strong number program using c?

Nikhil Aggarwal strong_number: The sum of factorials of digits of a number is equal to the original number. void strong_number() { int num,i,p,r,sum=0,save_num; printf("\nEn…ter a number"); scanf("%d",&num); save_num=num; while(num) { i=1,p=1; r=num%10; while(i

# What is strong number in C language?

Strong number is a number for which sum of factorials of the digits is equal to the given number. Eg: let us consider 145 where sum of factorials of the digits= 1!+4!+5!=1+2…4+120=145 i,e., the given number. c program for strong number is: #include #include void main() { int sof=0,n,dn,ctr,prod,rem; printf("Enter the number\n"); scanf("%d",&n); dn=n; while(n!=0) { prod=1,ctr=1; rem=n%10; while(ctr

# Factorial number's summation's program upto n numbers in c language?

include#includeint a,f,n,sum=0; printf("Enter any number"); scanf("%d",&n); f=1; for(a=1;a

# Write a program in C to find out Armstrong numbers?

include main() { int number, temp, digit1, digit2, digit3; printf("Printing all Armstrong numbers between 1 and 500:\n\n"); number = 001; while (number

# Write a c plus plus program to check a number is Armstrong number?

include #include // for std::pow() unsigned int get_length(unsigned int num,const unsigned int base=10) { unsigned int len=1; while(num && (num/=base)) ++len; return( le…n ); } bool is_armstrong(const unsigned int num,const unsigned int base=10) { unsigned int len=get_length(num,base); unsigned int sum=0; unsigned int tmp=num; while(tmp) { sum+=(unsigned int)std::pow((double)(tmp%base),(double)len); tmp/=base; } return(num==sum); } int main() { std::cout

# Write a program in c to find whether number is a Armstrong number or not?

include #include void main() { int a=1,b,y,m,q; while(a

# C program to find Armstrong number using for loop?

To find Armstrong numbers using a for loop, the loop simply generates decimal numbers in a given range and tests each of them to see if they are Armstrong numbers in a given b…ase. This is known as a brute force search. An Armstrong number is any number in base b with n digits such that the sum of each of the digits raised to the power of n is the number itself. Thus 1634 is an Armstrong number because it has 4 digits and (1^4) + (6^4) + (3^4) + (4^4) = 1 + 1296 + 81 + 256 = 1634. Armstrong numbers are dependant upon the base of the number. Thus the number 8 in base 3 is 22 3 , and 2^2 + 2^2 = 4 + 4 = 8. Therefore 22 3 is an Armstrong number (or, to put it another way, 8 is an Amstrong number in base 3). The first thing to determine is how many digits there are in a given number for a given base. For instance, the number 3 in base 10 (decimal) has only 1 digit (3), but in base 3 it has 2 digits (10 3 ). The number of digits is determined by continually dividing the decimal value of the number by the base until the result is zero (ignoring the remainder of each division). The number of divisions yields the number of digits in that base. Thus 17 10 requires 2 divisions: 17/10 = 1, 1/10 = 0, while 122 3 (17 decimal) requires 3 divisions: 17/3 = 5, 5/3 = 1, 1/3 = 0. Note that the division is always done on the base 10 value because the denominator (the base) is always a decimal value. The next problem is to determine the value of any digit at any position in a given number for a given base. The units position is always designated position 0 so for any position greater than 0, we divide the number by the given base raised to the power of the given position. This effectively moves the digit that we are interested in to the units position (position 0). We then divide this new number by the base and take the remainder, which is the digit we are actually looking for. Thus to find the digit in position 2 of the value 3412 decimal, we raise the base by a power of the position, thus 10^2 = 100. We then divide that into 3412 which yields 34 (ignoring the remainder). We then divide by the base and take the remainder. Thus 34/10 = 3 remainder 4. The digit 4 is therefore in position 2. Once we have these two functions in place, the rest is easy. Given a base, we simply loop through a sequence of numbers. On each iteration we determine the number of digits in the current number, then sum each of the digits in that number raised to the power of the number of digits. If the sum is the same as the number, we've found an Armstrong number and print it. Otherwise we move onto the next number. The following program will locate all the decimal Armstrong numbers in the range 0 through 268,435,455 (locating the first 29 of the 88 Armstrong numbers). You may change the base to any value in the range 2 through 16 to look at other bases. Note that in base 2, only the values 0 and 1 are Armstrong numbers. The PrintValue() function is an extra function for dealing with bases other than decimal. It simply prints any decimal value according to the given base. If you want to cater for other bases beyond 16, you must modify the switch statement to cater for all the symbols used by those bases, and adjust the if() statement at the beginning of the function to increase the range. I've also included a 64-bit unsigned integer definition to cater for huge numbers. If you increase the maximum to 0xffffffffffffffff the program will take many hours to complete. The larger the number, the more calculations that need to be done, thus the time to completion will rise exponentially. Back in 1985, it took about a year to calculate all the Armstrong numbers in base 16. Note that GetDigit() and GetDigitCount() have been inline expanded within the main() function to improve the overall processing speed. These functions are only actually used by the PrintValue() function when a non-decimal Armstrong number is discovered. The main function is verbosely commented for clarity. .
include typedef unsigned long UINT; typedef unsigned long long UINT64; UINT GetDigitCount( UINT64 n, UINT base = 10 ) { UINT count; count = 0; while( n ) { n /= base; ++count; } return( count ); } UINT GetDigit( UINT64 n, UINT pos, UINT base = 10 ) { UINT64 div; div = base; if( pos ) { while( --pos ) div *= base; n /= div; } return( n % base ); } void PrintValue( UINT64 n, UINT base = 10 ) { UINT digits, digit, pos; if( base < 2 || base > 16 ) return; if( n ) digits = GetDigitCount( n, base ); else digits = 1; for( pos = digits; pos; --pos ) { digit = GetDigit( n, pos-1, base ); if( digit < 10 ) printf( "%u", digit ); else { switch( digit ) { case( 10 ): printf( "a" ); break; case( 11 ): printf( "b" ); break; case( 12 ): printf( "c" ); break; case( 13 ): printf( "d" ); break; case( 14 ): printf( "e" ); break; case( 15 ): printf( "f" ); break; } } } } int main() { // Constants: const UINT64 max = 0xfffffff; // maximum possible value is 0xffffffffffffffff const UINT base = 10; // Local variables: UINT64 number, value, accumulator, digit_raised; UINT digit, digits, pos, raise, count; // Print the banner: printf( "Locating base %u Armstrong numbers in the range 0 through %I64u\n\n", base, max ); // Initialise the count: count = 0; // Loop through each value in the range. for( number = 0; number < max; ++number ) { // Store the value locally. value = number; // Determine the number of digits in the value. digits = 0; while( value ) { ++digits; // Reduce the value by an order of the base (ignore the remainder). value /= base; } // Restore the original value. value = number; // Initialise the accumulator. accumulator = 0; // Loop through each digit position while the accumulator is in range. for( pos = 0; pos < digits && accumulator < number; ++pos ) { // Extract the unit digit (right-most digit). digit = value % base; // Raise the unit digit by the power of the count of digits. digit_raised = digit; for( raise = 1; raise < digits; ++raise ) digit_raised *= digit; // Update the accumulator. accumulator += digit_raised; // Reduce the value by an order of the base (ignore the remainder). value /= base; } // Were all digits processed and does the accumulator equal the original number? if( pos digits && accumulator number ) { // Found an Armstrong number! printf( "Armstrong number %.2I64u: ", ++count ); if( base != 10 ) { PrintValue( number, base ); printf( " (%I64u)\n", number ); } else printf( "%I64u\n", number ); } } printf( "\n" ); return( 0 ); } Output: Locating base 10 Armstrong numbers in the range 0 through 268435455 Armstrong number 01: 0 Armstrong number 02: 1 Armstrong number 03: 2 Armstrong number 04: 3 Armstrong number 05: 4 Armstrong number 06: 5 Armstrong number 07: 6 Armstrong number 08: 7 Armstrong number 09: 8 Armstrong number 10: 9 Armstrong number 11: 153 Armstrong number 12: 370 Armstrong number 13: 371 Armstrong number 14: 407 Armstrong number 15: 1634 Armstrong number 16: 8208 Armstrong number 17: 9474 Armstrong number 18: 54748 Armstrong number 19: 92727 Armstrong number 20: 93084 Armstrong number 21: 548834 Armstrong number 22: 1741725 Armstrong number 23: 4210818 Armstrong number 24: 9800817 Armstrong number 25: 9926315 Armstrong number 26: 24678050 Armstrong number 27: 24678051 Armstrong number 28: 88593477 Armstrong number 29: 146511208

# Program in c to find whether number is a Armstrong or strong number?

include #include void main() { int number, sum = 0, rem = 0, cube = 0,temp; printf ("enter a number"); scanf("%d", &number); temp = number; while (number != 0) …{ rem = number % 10; cube = pow(rem, 3); sum = sum + cube; number = number / 10; } if (sum == temp) printf ("The given no is armstrong no"); else printf ("The given no is not a armstrongno"); }

Answered

In Computer Programming

# What is a perfect number in c language?

Perfect numbers have nothing to do with programming languages. Some of them are: 6, 28, 496, 8128, 33550336.

Answered

In C Programming

# How do you print number in digits in c language?

with printf

Answered

In Prime Numbers

# Which number is an Armstrong number?

0 137

Answered

In Math and Arithmetic

# How do you write the average n numbers program in c language?

Write your program and if you are having a problem post it here with a description of the problem you are having. What you are asking is for someone to do your homework for… you. no i am asking to verify my answer

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In C++ Programming

# What is programme in c plus plus of a Armstrong number?

include #include // for std::pow() unsigned int get_length(unsigned int num,const unsigned int base=10) { unsigned int len=1; while(num && (num/=base)) ++len; return( le…n ); } bool is_armstrong(const unsigned int num,const unsigned int base=10) { unsigned int len=get_length(num,base); unsigned int sum=0; unsigned int tmp=num; while(tmp) { sum+=(unsigned int)std::pow((double)(tmp%base),(double)len); tmp/=base; } return(num==sum); } int main() { std::cout