In number theory, the multiplicative digital root of a natural number n {\displaystyle n} in a given number base b {\displaystyle b} is found by multiplying the digits of n {\displaystyle n} together, then repeating this operation until only a single-digit remains, which is called the multiplicative digital root of n {\displaystyle n} . The multiplicative digital root for the first few positive integers are:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 4, 6, 8, 0, 2, 4, 6, 8, 0, 3, 6, 9, 2, 5, 8, 2, 8, 4, 0. (sequence A031347 in the OEIS) Multiplicative digital roots are the multiplicative equivalent of digital roots, with a major difference being that for natural numbers in base b = 10 {\displaystyle b=10} , the multiplicative digital roots can be 0 to 9, whereas digital roots can only be 1 to 9.
Definition Let n {\displaystyle n} be a natural number. We define the digit product for base b > 1 {\displaystyle b>1} F b : N → N {\displaystyle F_{b}:\mathbb {N} \rightarrow \mathbb {N} } to be the following:
F b ( n ) = ∏ i = 0 k − 1 d i {\displaystyle F_{b}(n)=\prod _{i=0}^{k-1}d_{i}}
where k = ⌊ log b n ⌋ + 1 {\displaystyle k=\lfloor \log _{b}{n}\rfloor +1} is the number of digits in the number in base b {\displaystyle b} , and
d i = n mod b i + 1 − n mod b i b i {\displaystyle d_{i}={\frac {n{\bmod {b^{i+1}}}-n{\bmod {b}}^{i}}{b^{i}}}}
is the value of each digit of the number. A natural number n {\displaystyle n} is a multiplicative digital root if it is a fixed point for F b {\displaystyle F_{b}} , which occurs if F b ( n ) = n {\displaystyle F_{b}(n)=n} . For example, in base b = 10 {\displaystyle b=10} , 0 is the multiplicative digital root of 9876, as
F 10 ( 9876 ) = ( 9 ) ( 8 ) ( 7 ) ( 6 ) = 3024 {\displaystyle F_{10}(9876)=(9)(8)(7)(6)=3024}
F 10 ( 3024 ) = ( 3 ) ( 0 ) ( 2 ) ( 4 ) = 0 {\displaystyle F_{10}(3024)=(3)(0)(2)(4)=0}
F 10 ( 0 ) = 0 {\displaystyle F_{10}(0)=0}
All natural numbers n {\displaystyle n} are preperiodic points for F b {\displaystyle F_{b}} , regardless of the base. This is because if n ≥ b {\displaystyle n\geq b} , then
n = ∑ i = 0 k − 1 d i b i {\displaystyle n=\sum _{i=0}^{k-1}d_{i}b^{i}}
and therefore
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