The simple trick with modulo 7 is take all the 3 bit sequences: (x & 111), (x & 111000) >> 3, (x & 111000000) >> 6, etc. and add them all together. It means you just have to do bitshift + add instead of all the complexity of division. Same applies to higher powers of 2^n-1 as well.

You can also combine those shift/adds by multiplying by a 1001001001 pattern or perhaps some 1000001000001 patterns.

Sure. That's why I said it was the same as the mod 9 tricks in base 10.

Yeah, sorry, there are extra steps here. If n= 43112609 and you have some 2n-bit number to reduce, you can split it into two n-bit words and simply add the words together, since 1 = 2^n (mod 2^n - 1). You can do this very quickly (ie you can add an arbitrary number of 2n -bit values together with only a small number of adds and shifts as overhead).

Yah-- as mentioned-- just like we can find the residue mod 9 by adding digits up in base 10.

There's notably also the all 1's case to deal with as an exception.