2013 May 23 06:56 PM MDT | Math7 Programming18 PHP4 Python4 3

What happens to the digital root as the numbers increase?

n dr(n)
0 0
1 1
2 2
3 3
9 9
10 1
11 2
12 3
18 9
19 1
 n | dr(n)
---+------
0 | 0
1 | 1
2 | 2
3 | 3
9 | 9
10 | 1
11 | 2
12 | 3
18 | 9
19 | 1


It looks like it is really just a cycle that goes from 1-9 over and over again (except 0). This can be represented using modulus.

$dr(x) = 1 + (x - 1) \bmod 9$

What happens if n < 0? Well, I've decided to just use the absolute value of it. That means that dr(-1) = 1.

Here is some PHP code:

<?php
for($i = -11;$i <= 20; ++$i) echo "dr($i) = ".(1 + (abs(\$i) - 1) % 9);


See that? There is no need for looping and recursion (nor iteration) to calculate the digital root. Why take O(≫ 1) time when you can take O(1) time?

Many languages handle negative modulus with truncation, but a special case must be created for some languages that use floored division:

# Python
def dr(i):
if i:
return 1 + (abs(i) - 1) % 9
return 0

for i in range(-11, 20 + 1):
print "dr(%d) = %d" % (i, dr(i))