$29 \cdot 1$ If a six-digit number is formed by repeating a certain three-digit number, such as 256256 or 678678, etc., any number of this type is always divisible by certain numbers. These numbers are
(A) only 7.
(B) only 11.
(C) only 13.
(D) 101.
(E) $7,11,13$.
(2nd American High School Mathematics Examination, 1951)
[Solution] Let the three-digit number that repeats in the six-digit number be $\mu$, then the six-digit number can be expressed as
$$
1000 \mu+\mu=\mu(1000+1)=1001 \mu .
$$

And
$$
1001=7 \cdot 11 \cdot 13 \text {. }
$$

Therefore, the answer is $(E)$.