Answer / venu

sol 1:
int fun(int i5DigitNum)
{
return 2<< i5DigitNum; // will overflow if the number is > 32
}

sol 2:
//assumption 32 bit machine
temp = i5DigitNum/32 + i5DigitNum%32 == 0 ? 0 :1 ;
char * temp2 = malloc(temp*4)
temp2[0] = 1 << i5DigitNum%32;

// now print this array as number!! :(

Answer / Raja Singh

Here is a simple C program that calculates 2 raised to the power of a 5-digit number without using big integers or exponential functions:n```n#include <stdio.h>n#define MAX 99999nint main() {n int num, result = 1;n printf("Enter a 5-digit number: ");n scanf("%d", &num);n if(num > MAX || num < 0){n printf("Invalid input. Please enter a 5-digit number.");n return 0;n }n for(int i = 1; i <= 5; i++){n result *= 2;n num /= 10;n result *= num;n }n printf("Result: %d", result);n return 0;n}

Answer / nitin katakdound

int main(int argc, char *argv)
{
long two_power, five_digit_number;

if(scanf("%ld",&five_digit_number)){
two_power = five_digit_number*five_digit_number;}
printf("\n 2 power of that number is %ld",two_power);

return 0;
}

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