When completed, the cross-number below will have one digit (from 0-9) in each cell, and no zeros in row (a) and in column (a).

         (a) (b) (c) 
            +---+---+
    (a)     |   |   |
        +---+---+---+
    (b) |   |   |   |
        +---+---+---+
    (c) |   |   |    
        +---+---+

ACROSS : (a) Abigail´s age.
         (b) Sum of Abigail´s age,
                    Blanche´s age,
                    Cynthia´s age, and
                    Darlene´s age.
         (c) Blanche´s age.

DOWN   : (a) Darlene´s age
         (b) Sum of three of the ages in (b) across.
         (c) Cynthia´s age.

Whose age was omitted from (b) down, and what are the ages of the 4 women ?

Note: this can be solved by hand. Those who will use the computer, give the others some time before posting the solution obtained this way. Tk you.

Solution:  Blanch's age was omitted.

Ages:

Abigail 16
Blanch  95
Cynthia 60
Darlene 29

Sum of 4 is 200
Sum of 3 is 105 (Blanch omitted)

The grid is:

     1   6

2    0   0

9    5

General procedure:

1.  Narrow the range of values for the sum of 4 and sum of 3.
2.  Examine possibilities for each age; narrow down to the one that is omitted
3.  Determine ages and sums

Specifics:

Assign variables as follows:

   J  K

L  M  N

P  R

So ages are as follows:

Abigail:  10J + K
Blanche:  10P + R
Cynthia:  10K + N
Darlene:  10L + N

Sum of 4 ages:  100L + 10M + N
Sum of 3 ages:  100J + 10M + R

1.  Begin with the observation that the max age for each is 99. 
So the absolute max sum of 4 ages is 99+99+99+99 = 396. So the max value of J = 3.
And since the sum of 4 ages is greater than the sum of 3 ages, max value of L = 3.
This means that the max age for Abigail and Darlene is 39.  Given that, the new max
sum of 4 ages is 39+99+99+39 = 276.  This refines the max value of J and L = 2.
And max age for Abigail and Darlene is 29.  Since row and column (a) cannot be zero,
then the possible combinations of J and L are J = L = 1, J = L = 2 and J = 1 and L = 2.
(Cannot have J = 2 and L = 1 because the sum of 3 cannot be larger than sum of 4).

2.  The basic relationship of the sums is:

(Sum of 4) = (Sum of 3) + (Omitted Age)

which is to say:

100L + 10M + N = 100J + 10M + R + (omitted Age, formulas as above).

Now we try each of the ages as the omitted age, and for each of the combinations of J and L,
and see what happens:

Abigail:  J = L = 1:  100 + 10M + N = 100 + 10M + R + 10 + K

Simlifying results in N = 10 + R + K.  This can't be a solution, because it requires N > 9.

Similarly, J = L = 2 gives the same result.

Finally, for J = 1, L = 2 we get (simplified) 90 = R + K - N.  This can't be a solition either
because it requires R or K to be > 9. 

So we can eliminate Abigail as the omitted age.

Using the same process, we can eliminate Cynthia and Darlene.

That leaves Blanch:  J = L = 1 results in a non-solution, as does J = L = 2.

J = 1, L = 2:  200 + 10M + N = 100 + 10M + R + 10P + R

Simplified:  100 + N = 10P + 2R, or 100 - 10P = 2R - N

Now, notice that with maximum value for R (9) and minimim for N (0),
the maximum value of 2R - N is 18.  So the only value that P can be is 9.

Now given P = 9, we get 10 = 2R - N.  The possibilities for R and N are:

N  R
0  5
2  6
4  7
6  8
8  9

Looks like Blanche is the one with the omitted age, but let's finish.

3.  Using the (sum of 4 ages) = sum of the 4 ages (substituting known
values J = 1, L = 2, P = 9):

200 + 10M + N = (10 + K) + (90 + R) + (20 + 9) + (10K + N)

Simplifying results in:  11K - 10M = 71 - R.

Now, with R being one of the values as above (5 thru 9) means the right
side of the equation will have values of 62, 63, 64, 65, or 66.
Now testing values of K (1 - 9) and M (0 to 9), we find that only one
combination of K and M results in an integer value for 11K - 10M.
Those values are K = 6, and M = 0, which gives R = 5, and therefore
N = 0.

So now we have all the variable values, sums and ages.

There are probably some logic shortcuts that I missed, and I look forward
to someone pointing them out.

Posted by dopey915 on 2005-10-27 13:56:29