"Double" Divisors

Note that $28 = 2 \times 14 = 4 \times 7 = 2 \times 2 \times 7$

If $\overline{abc}$ has only two distinct prime divisors, then its prime factorization is $p_1 \cdot p_2^{13}$ or $p_1^3 \cdot p_2^6$ . Where $p_1$ and $p_2$ are distinct primes. But $p_1 \cdot p_2^{13} \geq 2^{13} \cdot 3 > 1000$ and $p_1^3 \cdot p_2^6 \geq 3^3 \cdot 2^6 > 1000$ which are invalid.

Thus $\overline{abc}$ must have 3 distinct prime divisors. $28 = p_1 \cdot p_2 \cdot p_3^6$ for distinct primes $p_1, p_2, p_3$ . And because $p_1 \cdot p_2 \cdot p_3^6 \geq 2^6 \cdot 3 \cdot 5 = 960$ and the second minimum value of $p_1 \cdot p_2 \cdot p_3^6$ is $2^6 \cdot 3 \cdot 7 > 1000$ . So $\overline{abc} = 960$ only.

Therefore $960 = 2^6 \cdot 3 \cdot 5 \Rightarrow 960960 = 960 \cdot 1001 = 2^6 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ . $\overline{abcabc}$ has $(6+1)(1+1)^5 = \boxed{224}$ positive divisors.

abcabc=(1000abc)+abc=1001abc=7x11x13xabc. So we know that abc got 28 positive divisors and we have to multiply this number by the exponents+1 of 7, 11 and 13 obtaining: 28x(2x2x2)=224

if a number has prime factors of the form (a^x),(b^y),(c^z)......and so on then the number of factors are (x+1) (y+1) (z+1)

eg = 12 = (2^2) 3......therefore number of factors of 12 are (3 2)= 6

1) abcabc = 1001*abc

2) abcabc = 7 11 13*abc

3) 7 cannot be divisors of abc because if they were then the minimum value of abc would be abc = (2^6) * (3) * (7) = 1344 which is not a 3 digit number.

4) similary even 11and 13 cant be divisors of abc

5) therefore total divisors of abcabc are 28 + (2 2 2) = 224

Simple way:: 28 factors for abc. abcabc is 1001 x abc = 11 x 7 x 13 x abc. So Factors are 28 factors of abc + 28 of 7(abc) + 28 of 11(abc) + 28 of 13(abc) + + 28 of 7 x 11(abc) + 28 of 7 x 13(abc) + 28 of 13 x 11(abc) + 28 of 7 x 13x11(abc). That is 8 x 28 =224

By observation, the number of positive divisors to any number abcabc is 8 times the number of positive divisors to a number abc. So if abc has 28 divisors, abcabc has 28x8=224 divisors.

abcabc=1001 abc so number of divisors of abcabc is 28 2 2 2 as 1001 has 3 prime factors 7,11 and 13 each used once only.