Heres a quick funny (not so funny) question.....

Hey cilu :wave: you are not supposed to be speaking out such words here...
I don't understand why a software engineer should know about *clocks*, I think this is total peice of utter crap designed just to get some thing out of you which is almost surely required to be known.

To be frank I really go haywire when I see such questions, however I am not able to find out how answering/not answering such questions should decide my career.

Anyway, the OO solution in C++ is this:

enum PrecisionPerHour{Clock1 = 51, Clock2 = 60};

class IClock
{
const PrecisionPerHour precision_;
unsigned long time_;
unsigned long days_;
public:
IClock(PrecisionPerHour precision):precision_(precision), time_(0), days_(0){}
void Increment() {time_ += precision_; time_ %= 1440; if(time_ < (unsigned long)precision_) days_++;}
unsigned long GetTime() {return time_;}
unsigned long GetDays() {return days_;}
};

class FirstClock : public IClock
{
public:
FirstClock():IClock(Clock1){}
};

class SecondClock : public IClock
{
public:
SecondClock():IClock(Clock2){}
};

int _tmain(int argc, _TCHAR* argv[])
{
FirstClock c1;
SecondClock c2;

do
{
c1.Increment();
c2.Increment();
}while(c1.GetTime() != c2.GetTime());

std::cout << c1.GetDays() << ":" << c1.GetTime() / 60 << std::endl;

return 0;
}

Answer is: the clock shows the correct time again after 5 days and 16 hours.

And they did not list that answer.

AFAIK, an analog clock has no AM/PM indicator : And thus, a 12-hours shift is invisible.
Thus, the time will seem correct after 12*60/9==80 hours.
80 hours == 3 days + 8 hours.
Monday + 12 hours + 3 days + 8 hours == Thursday + 20 hours == Thursday at 8 hours PM.

Well, I don't see anything about an analog clock in the text of the problem.

A numeric clock shifting of 9 minutes per hour?

Well, you should stick to the original story and not make free assumptions.