A nice logic puzzle…

This puzzle was given to me in my first job interview…the initial time given was 3 minutes.  After 3 minutes I had still not solved it, so my interviewer asked, “If given some more time will you be able to solve it? Or you haven’t got any clue how to work it out?” …I said, “I am almost there”

Eventually I managed to solve it in a little over 4 minutes – and also got the job!

Here goes the puzzle:
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21200 is a peculiar number… 

If we number the digits from left to right from 0…i.e. First ‘2’ represents 0 th (zero) position, ‘1’ represents 1st position….till the last ‘0’ in the number represents 4th (fourth) position…
THEN, the number tells us: How many ZEROS there are in the zeroth position, how many ONES there are in FIRST position etc.

i.e. 21200 has ‘2’ zero’s, ‘1’ one’s, 2 two’s, ‘0’ three’s, and lastly ‘0’ four’s…or let me explain with headers for position and digits:

POSITION: 0 1 2 3 4
DIGITS in NUMBER: 2 1 2 0 0 …(== Actual number 21200)

Now the puzzle is:
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You have to find similar 8-digit number satisfying similar condition.

i.e. If we number the digits in that 8-digt number as 0 to 7 (starting from left), then each digit in that position should tell us how many times the number occurs in the entire number.

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I know, I may not have been able to clearly explain the problem statement; but hey…that is how the problem was explained to me as well….so I spent good time of the given 3-minutes to ‘understand’ the problem statement first!

So this is all you have got…check if you can solve. (The solution itself is not so difficult…once you figure out the approach)

~ Kaustubh
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