How to correctly convert days to a date with a fixed number of days in a month?

Question

PHP

Recently I have been training in OOP and found such a problem. I'm not sure about the correct solution, so I'm asking for help.
Good afternoon, there is a task of earth time converter. Somewhere in the universe, the planet NN dangles on which the day is 24 hours (like on earth), but there are 25 days in a month and 16 months in a year. You need to write a date converter between planet earth and planet NN. In the original, there are 5 such planets in the problem, but they differ only in the number of months in a year and the number of days in a month.
At the output, you need to get something like: Earth yyyy-mm-dd and NN: yyyy-mm-dd
It turns out that the earth day is the minimum indivisible unit (it is also a day on the planet NN)
In fact, I see here a class for the earth and a class for all other planets that will simply have their own getter and setter (to set the number of months in a year and days in a month)
There are 2 methods in the earth class

// Получаю количество дней от начала времен :)
  public function getAmountDaysByDate ($date) {
        $startDate = new DateTime("0001-01-01");
        $currentDate = new DateTime($date);
        $interval = $startDate->diff($currentDate);
        return $interval->days;
    }
  // Получаю дату по количеству дней от начала времен
    public function calculateDateByDays ($days) {
        $startDate = '0001-01-01';
        $date = new DateTime($startDate);
        $date->add(new DateInterval('P'.$days.'D'));
        $currentDate = $date->format('Y-m-d');
        return $currentDate;
    }

In general, everything is clear, there are 2 methods. The first returns the number of days between dates and the second returns the date by the number of days from the start.
But for the NN planet, the problem begins, maybe I don’t think something is right.

public $month = 16;
    public $days = 25;
    // Так как у меня большие проблемы с алгоритмами, лучше способа чем округлять дробь в меньшую сторону и отнимать от первоначального значения я не нашел... ( можно было использовать intdiv )
    public function getCurrentDateByDays($days)
    {
        $currentYear = floor ($days / ($this->month*$this->days));
        $daysMonth = $days - $currentYear*$this->month*$this->days;
        $currentMonth = floor ($daysMonth/$this->days);
        $currentDays = $daysMonth - $currentMonth*$this->days;
        return str_pad($currentYear+1, 4, '0', STR_PAD_LEFT).'-'.str_pad($currentMonth+1, 2, '0', STR_PAD_LEFT).'-'.str_pad($currentDays+1, 2, '0', STR_PAD_LEFT);
    }
 
   // А это наверняка можно решить более красиво чем отнимать 1
    public function getDaysByCurrentDate($date)
    {
        list($year, $month, $day) = explode("-", $date);
        $days = ($year - 1) * $this->month * $this->days + ($month - 1) * $this->days + $day;
        return $days;
    }

So that's what I was doing, when I counted the first few years in exel, the answers began to differ slightly (maybe problems with our earthly calendar, because xs how it is calculated in the depths of php for example 0002-01-01 and does it take into account the leap year and with what time period)
But rather I messed up everything in the getCurrentDateByDays method