Newsgroups: geometry.puzzles
From: haoyuep@aol.com (Dan Hoey)
Date: Wed, 24 Oct 2001 04:11:36 +0000 (UTC)
Subject: Re: Carts of Coal

Barry R. Clarke wrote:
> In Victorian London, there was a coal yard which customers would
> visit with their horse and cart. The yard sold two different size
> bags of coal, one being a two-digit number ab (i.e. 10a+b) times
> the weight of the other. One day, a trader turned up with his horse
> Neddy and asked for a number c of small bags and a number d of
> large bags to make the number e of cwt (hundred-weight) that his
> horse could pull. Shortly afterwards, another trader appeared with
> his horse Dobbin and asked for a number f of small bags and a
> number g of large bags to total the number h of cwt that Dobbin
> could pull. Each of a,b,c,d,e,f,g,h was a single digit, no two
> digits being equal and none of them being zero or one. What were
> the weights of the large and small bags?

If  s  is the weight of a small bag in cwt, we are given

  s(c + d(10a+b)) = e  and
  s(f + g(10a+b)) = h.

Solving both equations for  s,  we get

  e/(c + d(10a+b)) = h/(f + g(10a+b))

which can be rearranged to form

  ef-hc = (hd-eg)(10a+b) .

At this point, we could write a computer program to try all 40320
assignments of  {2,3,...,9}  to  {a,b,...,h},  but such a method
would be an anachronism in Victorian London.  So let us examine the
possibilities by hand.

Table 1 lists the products achievable by multiplying two different
numbers from {2,3,...,9} in order.

  Table 1 : products of two different numbers from {2,3,...,9}

   6 = 2x3        15 = 3x5        24 = 3x8 = 4x6  35 = 5x7   48 = 6x8
   8 = 2x4        16 = 2x8        27 = 3x9        36 = 4x9   54 = 6x9
  10 = 2x5        18 = 2x9 = 3x6  28 = 4x7        40 = 5x8   56 = 7x8
  12 = 2x6 = 3x4  20 = 4x5        30 = 5x6        42 = 6x7   63 = 7x9
  14 = 2x7        21 = 3x7        32 = 4x8        45 = 5x9   72 = 8x9

If the two-term expressions  ef-hc  and  hd-eg  were both zero, then
the products  ef=hc  and  hd=eg  would have to come from the three
duplicate factorizations 2x6=3x4, 2x9=3x6, and 3x8=4x6.  Then one of
the terms would be 3x6 or 4x6, and its two factors would appear in
separate terms of the other two-term expression, which is not
possible.

So none of the terms in  ef-hc = (hd-eg)(10a+b)  is zero.  10a+b  is
at least  10x2+3 = 23  and  |ef-hc|  is at most  8x9-2x3 = 66,  so
|hd-eg|  is either  1 or 2.  From examining nearby entries in Table 1,
we see this can arise in 14 ways, as listed in the first column of
Table 2.  The choices of  {h,d,e,g}  restrict our choices for {a,b},
so there is a new lower bound for  10a+b  in the second column.  The
third column has upper bounds for  |ef-hc|  (remembering that  ef-hc
and  hd-eg  have the same sign).  The eight ways eliminated by these
bounds are marked in the fourth column.

       Table 2 : small values for |hd-eg|

   |hd-eg|     min(10a+b)  max(|ef-hc|)

  3x4-2x5 = 2     67       9x5-6x3 = 27    no
  2x7-3x4 = 2     56       9x4-5x2 = 26    no
  3x5-2x7 = 1     46       9x7-4x3 = 51
  2x8-3x5 = 1     46       9x5-4x2 = 37    no
  3x6-2x8 = 2     45       9x8-4x3 = 60    no
  4x5-2x9 = 2     36       8x9-3x4 = 60    no
  4x5-3x6 = 2     27       9x6-2x4 = 46    no
  3x7-4x5 = 2     26       9x5-2x3 = 39    no
  4x7-3x9 = 1     25       8x9-2x4 = 64
  5x6-4x7 = 2     23       9x7-2x5 = 53
  4x8-5x6 = 2     23       9x6-2x4 = 46
  4x9-5x7 = 1     23       8x7-2x4 = 48
  6x7-5x8 = 2     23       9x8-2x6 = 60
  7x8-6x9 = 2     23       5x9-2x7 = 31    no

For the remaining six possibilities, we try all possibilities for
|ef-hc|  down to  min(10a+b) |hd-eg|,  and list  {a,b}  as the
remaining two available digits:

   |hd-eg|     min(|ef-hc|)     |ef-hc|     {a,b}

  3x5-2x7 = 1       46        9x7-4x3 = 51  {6,8}

  4x7-3x9 = 1       25        8x9-2x4 = 64  {5,6}
                              8x9-2x7 = 58  {5,6}
                              8x9-5x4 = 52  {2,6}
                              8x9-5x7 = 37  {2,6}
                              8x9-6x4 = 48  {2,5}
                              8x9-6x7 = 30  {2,5}
                              6x9-2x4 = 46  {5,8}
                              6x9-2x7 = 40  {5,8}
                              6x9-5x4 = 34  {2,8}
                              5x9-2x4 = 37  {6,8}
                              5x9-2x7 = 31  {6,8}

  5x6-4x7 = 2       46        9x7-2x5 = 53  {3,8}
                              9x7-2x6 = 51  {3,8}
                              9x7-3x5 = 48  {2,8}
                              8x7-2x5 = 46  {2,9}

  4x8-5x6 = 2       46        9x6-2x4 = 46  {3,7}

  4x9-5x7 = 1       23        8x7-2x4 = 48  {3,6}
                              8x7-2x9 = 38  {3,6}
                              8x5-2x4 = 32  {3,6}
                              8x7-3x4 = 44  {2,6}
                              8x7-3x9 = 29  {2,6}
                              8x5-3x4 = 28  {2,6}
                              8x7-6x4 = 32  {2,3} *
                              6x7-2x4 = 34  {3,8}
                              6x7-2x9 = 24  {3,8}
                              6x7-3x4 = 30  {2,8}

  6x7-5x8 = 2       46        9x8-2x6 = 60  {3,4}
                              9x8-2x7 = 58  {3,4}
                              9x8-3x6 = 54  {2,4}
                              9x8-3x7 = 51  {2,4}
                              9x8-4x6 = 48  {2,3}

The only solution is that  10a+b=32.  A small bag is 1/42 cwt and a
large bag is 16/21 cwt.  One horse pulled 7 cwt in 6 small bags and
9 large bags.  The other horse pulled 4 cwt in 8 small bags and
5 large bags.

Dan Hoey
haoyuep@aol.com