int and unsigned int

Westfw is right, I think. The actual printing method is this:

// Private Methods /////////////////////////////////////////////////////////////

size_t Print::printNumber(unsigned long n, uint8_t base) {
  char buf[8 * sizeof(long) + 1]; // Assumes 8-bit chars plus zero byte.
  char *str = &buf[sizeof(buf) - 1];

  *str = '\0';

  // prevent crash if called with base == 1
  if (base < 2) base = 10;

  do {
    unsigned long m = n;
    n /= base;
    char c = m - base * n;
    *--str = c < 10 ? c + '0' : c + 'A' - 10;
  } while(n);

  return write(str);
}

Other methods cast and muck around, but end up here. By sign extending into the unsigned long argument you will end up with extra "1" bits.