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Formatted question description: https://leetcode.ca/all/1837.html

# 1837. Sum of Digits in Base K

Easy

## 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

## Solution

Convert n into a base k number and calculate the sum of digits. Each time get a digit and add the digit to the sum. Finally, return the sum.

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

############

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

• // OJ: https://leetcode.com/problems/sum-of-digits-in-base-k/
// Time: O(log_K(N))
// Space: O(1)
class Solution {
public:
int sumBase(int n, int k) {
int ans = 0;
for (; n; n /= k) {
ans += 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

############

# 1837. Sum of Digits in Base K
# https://leetcode.com/problems/sum-of-digits-in-base-k

class Solution:
def sumBase(self, n: int, k: int) -> int:
res = 0

while n > 0:
res += n % k
n //= k

return res


• 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
}
}