# 1837. Sum of Digits in Base K

## Description

Given an integer n (in base 10) and a base k, return the sum of the digits of n after converting n from base 10 to base k.

After converting, each digit should be interpreted as a base 10 number, and the sum should be returned in base 10.

Example 1:

Input: n = 34, k = 6
Output: 9
Explanation: 34 (base 10) expressed in base 6 is 54. 5 + 4 = 9.


Example 2:

Input: n = 10, k = 10
Output: 1
Explanation: n is already in base 10. 1 + 0 = 1.


Constraints:

• 1 <= n <= 100
• 2 <= k <= 10

## Solutions

Solution 1: Mathematics

We divide $n$ by $k$ and take the remainder until it is $0$. The sum of the remainders gives the result.

The time complexity is $O(\log_{k}n)$, and the space complexity is $O(1)$.

• class Solution {
public int sumBase(int n, int k) {
int ans = 0;
while (n != 0) {
ans += n % k;
n /= k;
}
return ans;
}
}

• class Solution {
public:
int sumBase(int n, int k) {
int ans = 0;
while (n) {
ans += n % k;
n /= k;
}
return ans;
}
};

• class Solution:
def sumBase(self, n: int, k: int) -> int:
ans = 0
while n:
ans += n % k
n //= k
return ans


• func sumBase(n int, k int) (ans int) {
for n > 0 {
ans += n % k
n /= k
}
return
}

• function sumBase(n: number, k: number): number {
let ans = 0;
while (n) {
ans += n % k;
n = Math.floor(n / k);
}
return ans;
}


• /**
* @param {number} n
* @param {number} k
* @return {number}
*/
var sumBase = function (n, k) {
let ans = 0;
while (n) {
ans += n % k;
n = Math.floor(n / k);
}
return ans;
};


• impl Solution {
pub fn sum_base(mut n: i32, k: i32) -> i32 {
let mut ans = 0;
while n != 0 {
ans += n % k;
n /= k;
}
ans
}
}