Any integer composed of 3n identical digits, where n is a positive integer, is…

Concept: a number written as a single digit d repeated k times equals d × Rk, where Rk = (10k - 1)/9 is the repunit made of k ones. A divisor that has to work for every choice of the digit d cannot rely on d at all, so it must come from the repunit Rk; and repunits obey the splitting identity R3m = Rm × (1 + 10m + 102m).

Application to this question - write An for the repunit made of 3n ones, so the number described in the stem is N = d × An.

1. Claim: 3n divides An for every n ≥ 1. Prove it by induction on n.

2. Base case n = 1: A1 = 111 = 3 × 37, so 31 divides A1.

3. Inductive hypothesis: assume 3n divides An, and put m = 3n, so that An = Rm.

4. Splitting step: An+1 is the repunit of length 3m, and R3m = Rm × (1 + 10m + 102m), so An+1 = An × (1 + 10m + 102m).

5. The second factor 1 + 10m + 102m is a numeral made of three 1s separated by blocks of 0s, so its digit sum is 3 and it is divisible by 3.

6. Hence An+1 is divisible by 3n × 3 = 3n+1, which closes the induction.

7. Therefore N = d × An is a multiple of 3n for every repeated digit d and every n ≥ 1, whatever extra factors d may itself contribute.

Cross-check with concrete numbers:

• n = 2 gives 32 = 9 identical digits: 111111111 = 9 × 12345679 and 777777777 = 9 × 86419753, so 9 divides both, exactly as 3n predicts.

• That same 111111111 is odd, does not end in 0 or 5, and leaves remainder 6 on division by 7, so no positive power of 2, 5 or 7 divides it - none of those can be guaranteed for every digit d.

• The guarantee is exact and no stronger: 111111111 = 32 × 37 × 333667, so 33 = 27 does not divide it.

Result: an integer made up of 3n identical digits is always divisible by 3n.