Question

How to sum the digits of the number 2^N in JavaScript

Answer 1

function sumOfDigits(power) {
    const number = BigInt(2 ** power).toString(); // Handle large numbers
  
    return number.split('').reduce((sum, digit) => sum + Number(digit), 0);
}

console.log(sumOfDigits(15)); 
console.log(sumOfDigits(100)); 
console.log(sumOfDigits(1000)); 



/*
run:

26
115
1366

*/

 

70+ SQL courses for beginners and professionals

Answer 2

function sumOfDigits(power) {
    const number = BigInt(2 ** power).toString();
    let sum = 0;
    
    for (let digit of number) {
        sum += Number(digit);
    }
    
    return sum;
}

console.log(sumOfDigits(15)); 
console.log(sumOfDigits(100)); 
console.log(sumOfDigits(1000)); 



/*
run:

26
115
1366

*/

 

70+ SQL courses for beginners and professionals

Answer 3

function sumOfDigits(power) {
    const number = BigInt(2 ** power).toString();
    
    return Array.from(number).map(Number).reduce((sum, digit) => sum + digit, 0);
}

console.log(sumOfDigits(15)); 
console.log(sumOfDigits(100)); 
console.log(sumOfDigits(1000)); 



/*
run:

26
115
1366

*/

 

70+ SQL courses for beginners and professionals