Digital Root Calculator

The digital root of a positive integer is the single digit obtained by repeatedly summing the digits of the number. For example, dr(9875) = 9 + 8 + 7 + 5 = 29 → 2 + 9 = 11 → 1 + 1 = 2. The process always terminates in a single digit (1 through 9). By definition, dr(0) = 0. The digital root is also called the repeated digital sum or the iterated digit sum.

For positive integers: dr(n) = 1 + ((n − 1) mod 9), or equivalently, dr(n) = n mod 9 when n mod 9 ≠ 0, and dr(n) = 9 when n mod 9 = 0 (n > 0). Special case: dr(0) = 0. This formula gives the answer instantly without any iteration. Example: dr(9875) = 1 + (9875 − 1) mod 9 = 1 + (9874 mod 9) = 1 + 1 = 2.

Additive persistence is the number of times you must sum the digits before reaching a single digit. For example, 9875 → 29 → 11 → 2 takes 3 steps, so its additive persistence is 3. Most integers have persistence 1 or 2. The world record (as of 2025) for smallest number with persistence 11 is known; no number with persistence 12 has been found in base 10.

The digital sum is the result of summing the digits exactly once, which may itself be a multi-digit number. The digital root is the result of repeating the digit-sum process until only one digit remains. For 9875: digital sum = 9+8+7+5 = 29 (two digits), but digital root = 2 (single digit, after further reduction). For single-digit numbers, all three values coincide.

Casting out nines is an ancient arithmetic check: the digital root of n equals n mod 9. This means: (1) if digital root = 9, then 9 | n; (2) if digital root = 3 or 6, then 3 | n; (3) the digital roots of a sum, product, or difference equal the digital root of the result's digital root. This was used to verify long calculations before calculators - if the digital roots of inputs and output are inconsistent, there is an error.

Every digit d in position k contributes d × 10^k to the number. Since 10 ≡ 1 (mod 9), we have 10^k ≡ 1 (mod 9) for all k. Therefore n ≡ sum of its digits (mod 9). Repeating gives n ≡ digital root (mod 9). The only adjustment is for multiples of 9: their digit sum is also a multiple of 9, and the root is 9 (not 0), so the formula uses the 1 + (n−1) mod 9 form.

Applications include: (1) Divisibility checks: dr(n) = 9 means 9|n; dr(n) ∈ {3,6,9} means 3|n. (2) Arithmetic verification (casting out nines): check sums and products. (3) Cryptography and checksums: the ISBN check digit algorithm uses a similar modular sum. (4) Recreational mathematics: digital roots follow patterns in multiplication tables, Fibonacci numbers, and powers. (5) Computer science: hash functions sometimes use iterative digit sums.

The digital roots of multiples of any integer from 1 to 9 follow a repeating cycle of length 9. For the 7× table: 7, 14, 21, 28, 35, 42, 49, 56, 63 have digital roots 7, 5, 3, 1, 8, 6, 4, 2, 9 - the same nine values in a different order. For multiples of 9: the digital root is always 9 (since 9, 18, 27, ... are all multiples of 9).

In binary (base 2), summing bits gives the number of 1-bits (the Hamming weight or popcount). The digital root in base 2 is simply 0 or 1 - any binary number either has all zero bits (root 0) or eventually reduces to 1 (since summing bits gives a count ≥ 1 which keeps reducing). This equals n mod 1 = 0 or n mod (base−1) = n mod 1, adjusted to be the parity.

The digital roots of Fibonacci numbers in base 10 form a repeating cycle of length 24: 1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9,1,1,... (then repeats). This 24-cycle is directly related to the Pisano period of Fibonacci numbers modulo 9. Every ninth Fibonacci number has digital root 9.

Yes - the formula dr(n) = 1 + (n−1) mod 9 works for any positive integer, however large. For a 1000-digit number, you don't need to iterate: just compute the number mod 9 (or sum its digits once and compute that mod 9). This calculator handles standard JavaScript integers. For extremely large numbers, sum all digits manually then use the formula on the sum.