MHB Can You Crack the Divisibility by 7 Challenge in This Math Puzzle?

[Ackbach]

Here is this week's POTW:

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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 $d_{i}$ in $d_{1}d_{2}\dots d_{9}$ 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.]

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[Ackbach]

Re: Problem Of The Week # 214 - May 3, 2016

This was Problem A-3 in the 1995 William Lowell Putnam Mathematical Competition.

Congratulations to kaliprasad for his correct solution! It follows below:

Spoiler

We have the given number $N = \sum_{n=1}^{9} d_n * 10^{9-n} $.
Now replacing $d_i$ by $e_i$ we get $N + (e_i - d_i) * 10^{9-i} \equiv 0 \pmod 7 \cdots (1)$ for each i from 1 to 9.
Now let a new number be $\sum_{n=1}^{9} e_n * 10^{9-n}$;
replacing $e_i$ by $f_i$ we get $\sum_{n=1}^{9} e_n * 10^{9-n} + (f_i - e_i) * 10^{9-i} \equiv 0 \pmod 7\cdots(2).$
From (1)
$\sum_{i=1}^{9}( N + (e_i - d_i) * 10^{9-i}) \equiv 0 \pmod 7 $
or
$9N + \sum_{i=1}^{9}(e_i * 10^{9-i}) - \sum_{i=0}^{9}(d_i * 10^{9-i} ) \equiv 0 \pmod 7$
or $9N + \sum_{i=1}^{9}(e_i * 10^{9-i}) - N \equiv 0 \pmod 7 $
or $8N + \sum_{i=1}^{9}(e_i * 10^{9-i}) \equiv 0 \pmod 7 $
or $N + \sum_{i=1}^{9}(e_i * 10^{9-i}) \equiv 0 \pmod 7 \cdots (3)$
(1) , (2) and (3) hold for any specific $i$ as well. Add (1) and (2) and subtract (3) to get
$(f_i - d_i) * 10^{9-i} \equiv 0 \pmod 7$, and as $10^{9-i}$ is not divisible by 7 so $f_i- d_i$ and hence $d_i-f_i$ is divisible by 7.