Divisibility test for any number not a multiple of 2 or 5

A pattern I observed in 12th standard from the divisibility tests given for various numbers somewhere. It was easy to prove but I never made an effort before my 4th semester.

Here’s putting my observation into words:

“For any number x, if M is a multiple of the type (10p+9) then lets call p+1 ‘iterative multiplicand’ (IM) for x. Any number A is a multiple of x if the sum of IM times the units digit of A and the remaining number is also a multiple of x.”

Example: for 19, 2 is an IM. Now to check if 342 is a multiple of 19, we take sum of 2x2 (IM times the unit’s digit) and 34 (the remaining number) which is 38. Since 38 is a multiple of 19, so is 342.

The known divisibility tests for 3, 9, 11 can be derived from the same.

Proof: If M = 10p+9 is a multiple of x, then IM is p+1.
To check if A = a + 10b + 100c + 1000d + … is a multiple of x,
A = 10{(p+1)a + b + 10c + 100d + …} — 10ap — 9a
 = 10{(p+1)a + b + 10c + 100d + …} — aM
Hence if {(p+1)a + b + 10c + 100d + …} is a multiple of x (say N), A = 10N — aM is also a multiple of x.