Conference rusure::math

Take the 9 digit number as abcdefghi

e = 5  since 5|abcde

b,d,f,h are even since 2|ab ,4|abcd etc

6 | abcdef
-> 3 | a+b+c+d+e+f
since 3|a+b+c -> 3| d+e+f
since e = 5, 3|d+f+2 (=6,9,12,15,18,21,24..)
d,f are even
d+f+2 = 6,12,18
d+f = 4,10,16
{d,f} = {} (4)
        {2,8},{4,6} (10)
	{} (16)

d+e+f = 15  since d+f = 10

9|a+b+c+d+e+f+g+h+i
9|a+b+c+15+g+h+i
3|g+h+i

2|ab
4|abcd -> 4|ab*100+cd -> 4|cd
{c,d} = {1,6},{2,4},{2,8},{4,8}

c is odd gives c = 1 and d = 6 ...

d+e+f = 15 -> f = 4 ...

numbers left are {2,3,7,8,9}

8|abcdefgh -> 8|abcde*1000+400+gh
-> 8|gh
32,72,92,38,78,or 98

if 8|h -> 8|10*g but g is odd so
-> gh = {32,72,92}

-> h = 2

3|a+b+c -> 3|a+b+1 = (6,9,12,15,18,21,24)

a+b is odd
    a+b+1 = (6,12,18)
    a+b   = (5,11,17)
    {a,b} = {3,8} = 11
            {9,8}       = 17

    -> b = 8

So far we have :
a = {3,9}
b = 8
c = 1
d = 6
e = 5
f = 4
g = {3,7,9}
h = 2
i = {3,7,9}

?81654?2?
->
    3816547
    3816549 = 38165*100+7*7
    9816543
    9816547

Does 7 divide any of the above ?

    Now need the calculator...

and the answer is 

    7|3816547

so abcdefghi = 381654729

Adrian

    ... except that from this point
    
     >         ?81654?2?
    
    onwards you can put away your calculator and reason as follows:
    
    Up until now we've exhausted what the simple divisibility tests tell
    us and it just remains to get the division by 7 to yield some facts.
    
    Taking the partial remainders when you divide 7 into 10^6 we get the
    following test:
    
    	7 must divide (a + 5*b + 4*c + 6*d + 2*e + 3*f + g)
    
    and substituting for b to f this simplifies to (a+g) = 3 (mod 7)
    which, in the light of the previous restrictions means a=3 and g=7, so
    i=9.
    
    Dick
    
    

    This problem was in a local paper two weekends ago and the solution
    appeared last weekend. According to the article "... several readers
    pointed out that there were three solutions, this one, 921654387 and
    963258147".
    
    Well, several readers were mistaken. Both extra solutions obviously
    fail the 8 test (on 438 and 814).
    
    It's interesting that 987654321 almost works as well. It just fails the
    difficult 7 test.
    
    But the article did mention that the longest N-divisible number was ...
    Anyone care to have a go?
    
    (An N-divisible number's first N digits are divisible by N. The
    restriction on digits appearing exactly once is lifted.)
    
    Dick
    
    I had completely forgotten that I had already replied to this note. My
    memory is obviously on the way out. I found it by an Altavista search
    on the string "38165"!

3816547290608

I'll let someone else do the explaining if necessary...

>.4     
>   3816547290608
>
>   I'll let someone else do the explaining if necessary...
        
        I find
        
        	1575 N-Divisible 13-digit numbers
        	1132 N-Divisible 14-digit numbers
        	 770 N-Divisible 15-digit numbers
        	 571 N-Divisible 16-digit numbers
        	 335 N-Divisible 17-digit numbers
        	 180 N-Divisible 18-digit numbers
        	  90 N-Divisible 19-digit numbers
        	  44 N-Divisible 20-digit numbers
        	  18 N-Divisible 21-digit numbers
        	  12 N-Divisible 22-digit numbers
        	   6 N-Divisible 23-digit numbers
        	   3 N-Divisible 24-digit numbers
        		144408645048225636603816
        		360852885036840078603672
        		402852168072900828009216
        	   1 N-Divisible 25-digit number
        		3608528850368400786036725
        	   0 N-Divisible 26-digit numbers
        
        with the unique longest being 3608528850368400786036725.
        
        Dan

    re .4,.5
    
    These two notes seem to be in contradiction unless I have misunderstood
    (which wouldn't surprise).
    
    Is it the case that in dropping the condition that no digit may be
    repeated it is also the case that an N-divisible number of length 10
    (or longer) need not have the unique solution for length 9 (of the original
    problem) as a prefix?

    Derek
    
    The generalisation includes the fact that 0 can now be introduced as a
    digit. (.4) did not make use of that, (.5) did.
    
    (.5) agrees with what the article gave. Well that problem didn't last
    long, did it?
    
    Dan
    
    A pooter wasn't involved in that search by any chance?
    
    Dick

>    A pooter wasn't involved in that search by any chance?
        
        As I recall, one was. :-)
        
        Dan

The error came from reading the note and replies too quickly... I omitted to
notice that there isn't a unique N-divisible number of length 9 if the
restriction on the digits is lifted.

Thanks.