$29 \cdot 14$ when divided by $13511, 13903, 14589$, the largest divisor that can yield the same remainder is
(A) 28.
(B) 49.
(C) 98.
(D) an odd multiple of 7 greater than 49.
(E) an even multiple of 7 greater than 98.
(21st American High School Mathematics Examination, 1970)
［Solution］If $a, b, c$ when divided by $d$ all leave a remainder $r$, that is,
$$
a=a d+r, \quad b=\beta d+r, \quad c=\gamma d+r .
$$

Obviously, $a-b, b-c, a-c$ are all divisible by $d$.
Also, $(a-b)-(a-c)+(b-c)=0$, meaning that any common factor of two differences is also a factor of the third difference, and this holds true for the greatest common factor as well.
Given $13903-13511=7^{2} \cdot 2^{3}$, and $14589-13903=7^{3} \cdot 2$
and $\left(7^{2} \cdot 2^{3}, 7^{3} \cdot 2\right)=7^{2} \cdot 2$, where $(m, n)$ denotes the greatest common divisor (GCD) of $m$ and $n$.
Therefore, the answer is $(C)$.