Re: The NUMBER is 1687392

[Pham Thi Thanh Hong <smm68490@ait.ac.th>]

On Mon, 17 Mar 1997, Dung Trong Nguyen wrote:

> 
> Hi Hong,
> 
> I think your number is not divisible to 0.  No number can be.
> 
> This is mine (with a little help from my computer :))
> 
> 1) There no 0, so 9 digits remain.
> 
> 2) The least sinificant digit can't be odd number. If it is, the
> number can not be divisible to 2, 4, 6 and 8 and we don't have enough
> number.  Because there no 0 and the least significant digit is event,
> there no 5 in our number.
> 
> 3) 8 number 1,2,3,4,6,7,8 and 9 remain.
> 
> 4) If there no 9 in the number, 1+2+3+4+6+7+8 = 31, the number can not 
> be divisible to 3.
> 
> 5) If number include 9.  Because 1+2+3+4+6+7+8+9 = 40, there is only
> one way to make the 7-digit number be divisible to 9 is excepting 4
> from the set of digits.
> 
> 8) So 7 digits 1,2,3,6,7,8,9 remain to used.
> 
> 9) 1*2*3*6*7*8*9 = 18144
> 
> 10) We must check out 496 numbers those have the form (n * 18144) in
> the range (>=1016064 and <= 9997344) to get the valid and biggest
> number.  That number is 1687392, if the following program is bug-free
> :)
> 
> d~
> 
> --cut here--------------------------------------------------------
> 


Hi Du~ng,

Based on your assumptions that no 5 and no 0 digits included in X, I found
another number that is bigger than yours. It also satisfies the
conditions of problems. That is 9867312. This number is divisable for
every digits included such as 9,8,7,6,3,2, and 1.

I don't know what is the right X number for this problem because the
unclear assumption of every digits (whether these digits composed X or
every possible digits?)  

Ho^`ng.