@hoblos @phi Mensa Find a Ten-digit number containing each digit once, so that the number formed by the first n digits is divisible by n for each value of n between 1 and 10.

Question

@hoblos @phi Mensa  Find a Ten-digit number containing each digit once, so that the number formed by the first n digits is divisible by n for each value of n between 1 and 10.

anonymous · Answer

hehe i cheated and worked ahead before i posted it :P

kinggeorge · Answer

The obvious first step is to put 0 at the end so that it's divisible by 10, and then put 5 in the fifth place.

hoblos · Answer

yeah tahts visible
and we can know places of even and odd numbers
then i guess we have to try :P

anonymous · Answer

I think i got the answer, checking mine now and waiting for yours

kinggeorge · Answer

Let's be a little methodical about this then. Let's write our 9 digit number (I'm excluding the last place) as $$a_1\;a_2\;a_3\;a_4\;5\;a_6\;a_7\;a_8\;a_9$$Wher each $a_i$ is a digit. Then we know that $$2|a_2, \quad 4|a_3\;a_4, \quad 8|a_7\;a_8\;a_9, \quad 3|a_1+a_2+a_3 \quad 2|a_5\;a_6, \quad 3|a_1+a_2+a_3+a_4+a_4+a_5+a_6$$

hoblos · Answer

i got one....
3816547290

anonymous · Answer

You got it right @hoblos

kinggeorge · Answer

Thanks to google I have that too, but I'm still trying to analytically solve for it.

anonymous · Answer

and@ kinggeorge what do you mean i cant find it on google just tried.  shh no more good question medals xD

hoblos · Answer

btw i didnt google it!! 
just tried to arrange the numbers logically
like :
places 1,3,7,9  contain odd digits
places 2,4,6,8 contain even digits

sum of first three digits is divisible by 3
number formed by the third and the forth digits is divisible by 4......

kinggeorge · Answer

Fair enough.

phi · Answer

This problem was like doing a sudoku puzzle.  If I did not mess up, there is only the one solution 3816547290