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Relatively prime to their digital roots?
Inspired by Christine P's question about digital sums, what percentage of numbers are coprime to their digital roots?
The digital root of a number can be calculated by repeatedly taking the digital sum, until a single-digit number is reached.
For example, d(148) = d(1+4+8) = d(13) = d(1+3) = d(4) = 4.
Then gcd(148, 4) = 4.
1 Answer
- nudnik0Lv 610 years agoFavourite answer
The digital root d(n) of a number n is its remainder when divided by 9 (unless the number is a multiple of 9, in which case its root is 9.) So, which numbers n are coprime to d(n) can be determined by looking at the residue classes of n modulo 9 and modulo d(n).
If n ≡ 0, 3 or 6 (mod 9), 3 divides d(n), so n cannot be coprime to d(n).
If n ≡ 1 (mod 9), n is coprime to d(n).
If n ≡ 2, 4 or 8 (mod 9), n is coprime to d(n) iff n ≡ 1 (mod 2).
If n ≡ 5 (mod 9), n is coprime to d(n) iff n ≢ 0 (mod 5).
If n ≡ 7 (mod 9), n is coprime to d(n) iff n ≢ 0 (mod 7).
Adding up the fractions of the integers covered by these residue classes, the answer is that 1/9 + (1/2)*(3/9) + (4/5)*(1/9) + (6/7)*(1/9) = 97/210 = 46.19% of integers are coprime to their digital roots.
