Interchanging Digits!

Relevant wiki: Application of Divisibility Rules

It is easy to see, that if two integers are both divisible by 7, then their difference is also divisible by 7.

$\overline{abcd} - \overline {dbca} = 1000a +100b+10c+d-(1000d+100b+10c+a)=999(a-d)$

As 999 is not divisible by 7, therefore $7 \mid (a-d)$ .

Now, we have the following cases:

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I. $a - d = 0 \iff a=d$

As $7 \mid 1001$ ( and so are the multiples of 1001), therefore we have a solution if and only if the (not necessarily two digit, e.g. 00 or 07) number formed of the middle two digits ( $\overline {bc}$ ) is divisible by 7.

We have 15 solutions ( $\overline {bc} = 7k, k \in \mathbb {Z} , 0 ≤ k ≤ 14$ ) per value of digit a, and

9 × 15 = 135 solutions altogether in this case.

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II. a - d = ±7

As $a \in$ {1,2,3,4,5,6,7,8,9} , the possible (a,d) values are (1,8), (8,1), (2,9), (9,2) and (7,0).

Since 1008, 8001, 2009, 9002 and 7000 are all divisible by 7, we have a solution (similarly to case I.) iff $7 \mid \overline {bc}$

This means, that we have 15 solutions per pair of (a,d) values and

5 × 15 = 75 solutions altogether in this case.

$\\$ Therefore, the total number of solutions is:

$135 + 75 = \boxed {210}$