# The digital root of factorials from 6! to infinity! is always 9.

While observing digital roots of factorials, i observed that ,

digital root of (5+n)! where 'n' is any natural number , is always 9.

The reason lies in the number 720. It can also be written as 6! . And digitalroot of (6!× n) where 'n' is always a natural number, is always 9. And thus the statement holds true for factorials after 6! . Has it been observed before? Does it's proof exist? I am stuck at writting it's proof using mathematical induction. Can anybody help?

• The digital root of any number divisible by 9 is 9. Have you seen the divisibility test for 9? Commented Sep 10, 2015 at 19:31
• Yes just stuck at writting it mathematically. Commented Sep 10, 2015 at 19:42
• $6!=1*2*3*4*5*6=1*2*4*5*2*3*3=1*2*4*5*2*9$ (divisible by $9$) Commented Apr 18, 2017 at 13:44

The factorial numbers greater than 5! are all divisible by 9, as we have 3 twice in their expansion. The divisibility test for 9 is that the digital root is 9.

The proof for the divisibility rule for 9 is essentially the same as the proof for the divisibility rule for 3.

For any integer x written as an· · · a3a2 a1a0 we will prove that if 9|(a0 + a1+ a2+ a3 ... + an), then 9|x and vice versa.

First, we can state that

x = a0 + a1×10 + a2×102 + a3×103... + an×10n

Next if we let s be the sum of its digits then

s = a0 + a1 + a2 + a3 + ... + an .

So

x - s = (a0 - a0) + (a1 × 10 - a1) + (a2×102 - a2) + ... + (an×10n - an)

                           = a1(10 - 1) + a2(102 - 1) + ... + an(10n - 1).


If we let bk = 10k - 1, then bk = 9...9 (9 occurs k times) and bk ­=9(1…1) and we can rewrite the previous equation as

x - s = a1(b1)+ a2(b2)+ ... + an (bn)

It follows that all numbers bk are divisible by 9, so the numbers ak×bk are also divisible by 9. Therefore, the sum of all the numbers ak×bk (which is x-s) is also divisible by 9.

Since x-s is divisible by 9, if x is divisible by 9, then so is s and vice versa.