Another way of checking divisibility by 23

If only for the purpose of the exercise, we can now examine divisibility by 23, actually divisibility by 92, in the same way we examined divisibility by 91 (see “A divisibility rule for both 7 and 13” below). We convert base 100 to base 101 so we can work with 92 instead of 91. 

Example:

We have seen from the previous post 5 30 57 base 100 to be 5 20 32 base 101. 5 20 32 = 50 + 20 - 5  32 = 65 32 = 65 * 10 + 3  2 - 65. But remember our base of operations is 101, so 65 * 10 = 644.  644 + 30 + 2 - 65 = 611 = 71 - 6 = 65 = 19 (mod 23).

Another example:

234871 (100) = 23  48 - 2*23  71 - 23  =  23  02  48 =  23  02  48 - 2 = 23 02 46 (101)

23 02 46 = 23*10 + 2 - 23  46  =  228 (101) + 2 - 23  46 = 207 (101)  46  =  25 46  =  25*10  +  46  - 25  =  248 (101)  +  46  - 25  =  269  =  87

87 leaves a remainder of 18 when divided by 23 so 234871 (mod 23) = 18