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.*
Good work dopey915. My solution uses the letters A through G to represent the digits in the grid:

A B

C D E

F G

I started out as you did. Because none of the four ages could be greater than 99, the sum of their ages, CDE, could be no more than 396. This makes 3 the maximum possible value of C.

A B Abigails age

F G Blanch's age

B E Cynthia's age

__+ C F Darlene's age__

C D E

We also know that the sum of three of the four ages is ADG, so A is less than or equal to C. Let's use XY to represent the missing age. Then we have:

A D G

__+ X Y__

C D E

Because X cannot be 0, and both D and X are no greater than 9, the D in the center of CDE can result only if X is 9 and G+Y =E+10 so that a one is carried (yielding D + X + 1 = D +10.) This also means that another carry is required, yielding A+1=C.

Our original upperbound on the value of C was 3, but now we can replace it with the value 2 (because the sum of the four ages must now be less than 266), and know that A = C-1=1. Since X represents the tens digit of one of four women's ages, we know that the missing age belongs to Blanche or Cynthia, because Abigails age is 10 + B, and Darlene's age is 20 + F. But, we know that Cynthia's age can not be the missing one, because then XY=BE yielding:

A D G 1 D G

__+ B E __ => __+ 9 E__

C D E 2 D E

which isn't possible (G can't be 10, and because a carry is required, G can't be 0.) **So Blanche's age is the missing one** and we have, instead:

A D G 1 D G

__+ F G __ => __+ 9 G__

C D E 2 D E

We also have, the sum of the four ages:

1 B

9 G

B E

__+ 2 9__

2 D E

So, from the latter, B+G=1 or B+G=11. B is the 10's digit of Cynthia's age, so it can't be 0. And B = 1 isn't large enough to produce the carry digit's required to make the sum of the four ages greater than 199. So, B+G=11. This means a 2 will be carried to the tens column and B must be at least 6 to achieve this. And, from the equation involving the missing age (ADG+FG=CDE) we know that G+G=10+E, meaning G can be no smaller than 5. This leaves us with only one solution for B+G=11. B =6 and G=5. Plugging these values for B and G into our two sum equations yields E=0 and D=0. So we now have all the ages:

**16 Abigail's age**

** 95 Blanche's age**

** 60 Cynthia's age**

**+29 Darlene's age**

205