x * 10 = x * 8 + x * 2 = (x << 3) + (x << 1)

int divide10(int32_t num)
{
    return num / 10;
}

mov      eax, 1717986919      ; 66666667H
imul     ecx
sar      edx, 2

mov      eax, edx
shr      eax, 31
add      eax, edx

int r = (int)((x * 0x66666667L) >> 34);
r += (r < 0);

(x * 0x66666667L) >> 34 == [(10 * q + r) * (0x66666667L) / 2^34]

(2^34 + 6)/10 * (10 * q + r) = q * 2^34 + 6 * q + 0x66666667L * r

6 * q + 0x66666667L * r < 6 * 214748364 + 0x66666667L * 9 == 15676630635 < 2^34

// http://www.hackersdelight.org/divcMore.pdf
uint32_t divu10(uint32_t n, uint32_t *rem) {
  uint32_t q, r;
  q = (n >> 1) + (n >> 2);
  q = q + (q >> 4);
  q = q + (q >> 8);
  q = q + (q >> 16);
  q = q >> 3;
  //  orig: r = n - q*10;
  r = n - ((q << 3) + (q << 1));
  *rem = r > 9 ? r - 10 : r;

  //  orig: return q + ((r + 6) >> 4);
  return q + (r > 9);
}


uint64_t divu64_10(uint64_t n, uint32_t *rem) {
  uint64_t q, r;
  q = (n >> 1) + (n >> 2);
  q = q + (q >> 4);
  q = q + (q >> 8);
  q = q + (q >> 16);
  q = q + (q >> 32);
  q = q >> 3;
  //  orig: r = n - q*10;
  r = n - ((q << 3) + (q << 1));
  *rem = r > 9 ? r - 10 : r;

  //  orig: return q + ((r + 6) >> 4);
  return q + (r > 9);
}