(42, all flavours) Integer Division - how?

[Albert Chan]

Since c ≠ 0, or 5 (mod 10), we can rewrite the equality as:

0 ≡ (2*Qq ± 1) * (2c) (mod 10), where c = last non-zero decimal digit of Rr
0 ≡ (2*Qq ± 1) * c (mod 5)

+1 case: Qq ≡ 2 (mod 5)
−1 case: Qq ≡ 3 (mod 5)

But, -1 case should return Qq-1.
This meant half-way case floor-divide quotient always in 5n+2 form

PHP Code:

function idiv7b(Cc00, Xx)
    local Qq, Rr = d2(Cc00/Xx), Cc00%Xx
    local c = floor(Qq)
    if Qq ~= c then return c end
    if Rr < Xx-Rr then return c end     -- Cc00/Xx rounded-down to c
    if Rr > Xx-Rr then return c-1 end   -- Cc00/Xx rounded-up to c
    return c - (c-2)%5                  -- halfway case
end