Digital root
Encyclopedia : D : DI : DIG : Digital root
The digital root (also Repeated digital sum) of a number is the number received by adding all the digits, then adding the digits of that number, and then continuing until a single-digit number is reached.
For example, the digital root of 65,536 is 7, because [6+5+5+3+6 = 25] and [2+5 = 7]
Special cases of digital roots of particular numbers are:
- Digital root of a square is 1, 4, 7, or 9
- Digital root of a perfect cube is 1, 8 or 9
- Digital root of a prime number (except 3) is 1, 2, 4, 5, 7, or 8
- Digital root of a power of 2 is 1, 2, 4, 5, 7, or 8
- Digital root of a perfect number (except 6) is 1
- Digital root of a star number is 1 or 4
- Digital root of a triangular number is 1, 3, 6 or 9
- Digital root of a factorial ≥ 6! is 9.
Digital roots can be used as a sort of checksum. For example, since the digital root of a sum is always equal to the digital root of the sum of each summand's digital root, somebody adding long columns of large numbers will often find it reassuring to apply casting out nines to his or her result — knowing that this technique will catch the majority of errors.
Digital roots are used in Western numerology, but certain numbers deemed to have occult significance (such as 11 and 22) are not always completely reduced to a single digit.
Formal definition
Let [f(n)] denote the sum of the digits of [n]. Eventually the sequence [f(n),f(f(n)),f(f(f(n))),\dotsb] becomes constant. Let [f_(n)] (the digital sum of [n]) represent this constant value.Example
Let us find the digital sum of [1853].
- [f(1853)=17\,]
- [f(17)=8\,]
Proof that a constant value exists
But how do we even know that the sequence [f(n),f(f(n)),f(f(f(n))),\dotsb] eventually becomes constant? Here's a proof:
Let [x=d_1+10d_2+\dotsb+10^d_n], with [0\le d_i\in\mathbb<10] (For all [i], [d_i] is an integer greater than or equal to [0] and less than [10]). Then, [f(x)=d_1+d_2+\dotsb+d_n]. This means that [f(x)
The formula is:
To generalize the concept of digital roots to other bases b, one can simply change the 9 in the formula to b - 1.
From Wikipedia, the Free Encyclopedia. Original article here. Support Wikipedia by contributing or donating.Ramans' Formula
[ \mbox(n) = \begin n\ (\ 9)\ n\ \ne 0\ (\ 9) \ 9\ \ \ \ \ \ \ \ \ \ \ \ \ n\ \equiv 0\ (\ 9) \end ]
[ \mbox(n) = 1\ +\ [n-1 ( 9)] ]
See also
External links
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