37289 -> subtract 49 -> 37240 -> divide by 10 -> 3724 -> subtract 14 -> 3710 -> 371 -> subtract 21 -> 350 -> this is 35 * 10, and 35 is a multiple of 7.
The method described in the article is equivalent to my method, with the stipulation that one always subtracts, and that there is a mechanical process for picking the number to subtract (i.e. you multiply the number by 21).
37289 -> drop 9 from the end and subtract 18, which is equivalent to subtracting 189 and dividing by 10 -> 37289 - 189 / 10 -> 37100 / 10 -> 3710 -> drop 0 from end and subtract 0 -> 371 -> drop 1 from end, subtract 2, equivalent to subtracting 21 and dividing by 10 -> 371 - 21 / 10 -> 350 / 10 -> 35.
Advantages of my method:
a) This works with divisibility by any number at all (except for the factors of 2 and 5, which can be dealt with easily). For example, divisibility by 13:
594875 -> subtract 65 -> 594810 -> divide by 10 -> 59481 -> add 39 -> 59520 -> 5952 -> subtract 52 -> 5900 -> 59 -> not multiple of 13.
b) You don't need to remember any magic number to do it--in this case, you take the last digit, multiply it by the magic number -2, and add it to the remainder of the number. (For divisibility by a general m (coprime to 10), a magic number is any x such that x * 10 ≡ 1 (mod m). We see that adding 5 times the last digit would also work for divisibility by 7.)
c) It is quite elementary. It's based on the facts that 1) adding a multiple of 7 (or whatever your number is) won't change the divisibility, and 2) dividing by 10 won't change the divisibility.