# Scala algorithm: Compute single-digit sum of digits

Published

## Algorithm goal

\(909\) sums to \(18\), which sums to \(9\).

Find the best way to compute it for any positive \(n\), efficiently.

## Test cases in Scala

```
assert(sumOfDigits(0) == 0)
assert(sumOfDigits(2) == 2)
assert(sumOfDigits(8) == 8)
assert(sumOfDigits(10) == 1)
assert(sumOfDigits(16) == 7)
assert(sumOfDigits(32) == 5)
assert(sumOfDigits(64) == 1)
assert(sumOfDigits(101) == 2)
assert(sumOfDigits(109) == 1)
assert(sumOfDigits(128) == 2)
assert(sumOfDigits(256) == 4)
assert(sumOfDigits(512) == 8)
assert(sumOfDigits(999) == 9)
assert(sumOfDigits(1024) == 7)
assert(sumOfDigits(2048) == 5)
assert(sumOfDigits(4096) == 1)
```

## Algorithm in Scala

4 lines of Scala (version 2.13), showing how concise Scala can be!

## Explanation

This involves a bit of mathematics to figure out - but we can get an \(O(1)\) solution here:

The sum of digits of sum of digits... is equal to that digit modulo 9; if we take a number that is composed of digits\(abcd\), which equals \(10^3 \times a + 10 ^ 2 \times b + 10 \times c + d\), and then using %/modulo 9 we get:\(a + b + c + d (\mod 9)\), which is also \(((a + b) (\mod 9) + (c + d) (\mod 9))(\mod 9)\). Combining with a modulo table, you will notice thatindeed for example \(8 + 7 = 15\), and applying modulo 9 on \(15\) gives us \(6\), which is precisely the sum of digits \(1\) and \(5\). (this is Â© from www.scala-algorithms.com)