1995 Putnam A3

The number $d_{1}d_{2}\dots d_{9}$ has nine (not

necessarily distinct) decimal digits. The number $e_{1}e_{2}\dots

e_{9}$ is such that each of the nine 9-digit numbers formed by

replacing just one of the digits $d_{i}$ is $d_{1}d_{2}\dots d_{9}$

by the corresponding digit $e_{i}$ ($1 \leq i \leq 9$) is divisible

by 7. The number $f_{1}f_{2}\dots f_{9}$ is related to

$e_{1}e_{2}\dots e_{9}$ is the same way: that is, each of the nine

numbers formed by replacing one of the $e_{i}$ by the corresponding

$f_{i}$ is divisible by 7. Show that, for each $i$, $d_{i}-f_{i}$ is

divisible by 7. [For example, if $d_{1}d_{2}\dots d_{9} = 199501996$,

then $e_{6}$ may be 2 or 9, since $199502996$ and $199509996$ are

multiples of 7.]

数字 $d_{1}d_{2}\dots d_{9}$ 有 9 个(不是

必须不同)十进制数字。数字 $e_{1}e_{2}\dots

e_{9}$ 使得由以下组成的九个 9 位数字中的每一个

仅替换其中一位数字 $d_{i}$ 为 $d_{1}d_{2}\dots d_{9}$

被相应的数字 $e_{i}$ ($1 \leq i \leq 9$) 整除

7. 数字 $f_{1}f_{2}\dots f_{9}$ 与

$e_{1}e_{2}\dots e_{9}$ 的方式相同：即九个中的每一个

通过将 $e_{i}$ 之一替换为相应的数字而形成的数字

$f_{i}$ 能被 7 整除。证明，对于每个 $i$，$d_{i}-f_{i}$ 是

能被 7 整除。 [例如，如果 $d_{1}d_{2}\dots d_{9} = 199501996$，

那么 $e_{6}$ 可能是 2 或 9，因为 $199502996$ 和 $199509996$ 是

7的倍数。]