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
my solution
