Conference rusure::math

It's easy with a computer...
I searched candidates N that were just below  K * 10^10, such that
(N mod 7*13*9) = 7*13, and found several pretty quickly.

Are there any more shortcuts to doing this by hand?

How about N = 91...91, with 91 repetitions of '91'?

N = 7*13*101...01
sum of digits of N = 10*91 = 7*13*10.

        7*13 = 91 so you want both N and the sum of the digits of N
        to be multiples of 91.  7*11*13 = 1001 so if you write the
        digits of N as say abc,def,ghi then 91 divides N if and only
        if 91 divides abc - def + ghi.  So write down a number the
        digits of which add to 91 while keeping the "alternating
        series" of thousands a multiple of 91.  It is easy to pair
        up abc,abc which has an even total; the digits of 91 also
        add up to even but the digits of 182 = 2*91 add up to even.
        So try numbers 182,abc,abc,def,def where the digits add up
        to 91 (so a + b + c + d + e + f = (91 - 1 - 8 - 2)/2 = 40).
        40 = 27 + 13 so
        
        	have/need	999,999		553,553		182
        	  0 / 91        54 / 37         80 / 11        11/0
        
        So all of these will do for N:
        
        	182,553,553,999,999
        	182,999,999,553,553
        	553,553,999,999,182
        	999,999,553,553,182
        
        etc.
        
        Dan
        

To make the problem interesting, I suggest we change it to finding the smallest
such N.  I offer N = 198,289,999,999 as a candidate.

Easy lemma.  The smallest such N has the sum of its digits = 91.

I just found N = 59,878,999,999.

Another easy lemma: the minimum N has 11 digits, none 0, and ends in 99999. 
This comes from the fact that 10^6-1 is divisible by 91, so you could "move" a
value of 1 to the right 6 places if the end weren't 99999.

The value in .5, N = 59,878,999,999, is minimal.

Sketch of proof: consider numbers with 11 digits whose initial digit is 1 to 5
and whose digit sum is 91.  Any such number can be reached in k-1 steps (k being
the initial digit) from the starting value 19,999,999,999, where a "step"
consists of moving 1 unit of value from some other decimal place into the
initial digit.  Furthermore, the N we want has 99999 as the last 5 digits, so we
can ignore transfers from those last 5 places.

The starting point 19,999,999,999 == 70 (mod 91), and the values of the relevant
powers of 10 modulo 91 are 10^10 == -10, 10^9 == -1, 10^8 == 9, 10^7 == 10, 10^6
== 1, and 10^5 == -9.  One step changes N mod 91 by (-10-v), where v is one of
-1, 9, 10, 1, or -9.  So we must have one of

  60-v1            == 0 (if initial digit is 2)
  50-(v1+v2)       == 0 (if 3)
  40-(v1+v2+v3)    == 0 (if 4)
  30-(v1+v2+v3+v4) == 0 (if 5)

The first three are impossible, and the last can only work if the values v1..v4
are 10,10,9,1 in some order.  So the value is 59,999,999,999-(10^8+2*10^7+10^6)
= 59,878,999,999.

    The published solutions parallel the preceding notes, including
    observing the repetition from 1001 and the search for the smallest such
    number.  Additional solutions include 2002 repeated to get digit sum
    364, 546 1s, 91 91s, 7 364s, 78 7s, and 7^25*13.
    
    
    				-- edp