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.
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Submitted by pcbouhid
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Rating: 3.0000 (3 votes)
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Solution:
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(Hide)
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Let´s identify the cells: (a) (b) (c)
+---+---+
(a) | A | B |
+---+---+---+
(b) | C | D | E |
+---+---+---+
(c) | F | G |
+---+---+There are 4 possible cases (the name between parentheses is the name of the woman whose age was omitted):
CASE I CASE II CASE III CASE IV
(Abigail) (Cynthia) (Darlene) (Blanche)
ADG ADG ADG ADG
+ AB + BE + CF + FG
---- --- --- ---
CDE CDE CDE CDECASE I: In the middle column, A is 9 or 0, and since A is in row (a) where there is no zero, A = 9. Then C is more than 9 in the sum. So this case is impossible.
CASE II: B is 9 or 0, so B = 9 (B is in row (a)). Then 1 was carried from the right column. But G must be 0, so 1 could not be carried from the right column. This situation is impossible. So Case II is not a correct case.
CASE III: In the middle column, C is 9 or 0. Then, C = 9. But AB + BE + CF + FG = CDE. So C in CDE could not be greater than 3. This situation is impossible. So Case III is not a correct case, too.
CASE IV: This must be the correct case, and Blanche´age was omitted from (b) down.
F is 0 or 9, so F = 9. Then 1 was carried from the right column and 1 was carried from the middle column. Because 1 was carried from the right column, G is greater than 4. Because 1 was carried from the middle column, C is 1 more than A.
Because this is the correct case and once AB + BE + CF = ADG, then A = 1 and C = 2, or A = 2 and C = 3.
But, if A = 2 and C = 3 in the middle column, then A cannot be 2 in ADG. So A = 1 and C = 2.
Because F = 9 and G is greater than 4, (B + F + G) in the right column of AB + BE + CF + FG must be 20 (B + F + G must end in 0).
Then, because F = 9, (B + G) = 11. Because (B + G) = 11 and G is greater than 4, B is 6 or less.
Because A = 1 and C = 2 and F = 9, (A + B + C + F) is at least 18; so B is 6 or more. Thus B = 6. So G = 5.
D = 0 in AB + BE + CF + FG = CDE, and E = 0 in ADG + FG = CDE.
The complete grid is: (a) (b) (c)
+---+---+
(a) | 1 | 6 |
+---+---+---+
(b) | 2 | 0 | 0 |
+---+---+---+
(c) | 9 | 5 |
+---+---+The ages: Abigail - 16, Blanche - 95, Cynthia - 60, and Darlene - 29.
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