Grand Quiz Question #13
The 13th Question of the #-Link-Snipped-# is as follows -
[Worth 15 CE Coins]
Q. Fill numbers from 0 to 9 in the boxes below.

Rules are:
1. None of the digits can be repeated in the boxes.
2. The boxes that have letters a, b, c and d can't have the digit '0'
Information Given -
a (across) - W's age.
a (down) - Sum of the three ages in c - across
b (down) - Y's age
c (across) - Sum of ages of W, X, Y, Z
c (down) - Z's age
d (across) - X's age
Show the numbers in the boxes and tell us whose age among, W, X, Y and Z was not written in a down?
And yeah, tell us your approach in finding the solution!
---------------------------------------------------------------
Fastest to answer correctly wins!
[Worth 15 CE Coins]
Q. Fill numbers from 0 to 9 in the boxes below.

Rules are:
1. None of the digits can be repeated in the boxes.
2. The boxes that have letters a, b, c and d can't have the digit '0'
Information Given -
a (across) - W's age.
a (down) - Sum of the three ages in c - across
b (down) - Y's age
c (across) - Sum of ages of W, X, Y, Z
c (down) - Z's age
d (across) - X's age
Show the numbers in the boxes and tell us whose age among, W, X, Y and Z was not written in a down?
And yeah, tell us your approach in finding the solution!
---------------------------------------------------------------
Fastest to answer correctly wins!
Replies
-
Ambarish GaneshLargest possible age: 98
2nd largest : 87
then 76, 65, 54, 43, 32, 21, 10. 10 is the least possible age.
Trying 87 and 76, we get the other ages as 23 and 30, with 1 in the middle box.
Left the blue ones out!
The ages of X and W were not written in a down!(from question) ๐ -
Gandalf-----------
| 4 | 1 | 6 |
| 2 | 0 | 3 |
| 9 | 5 | 7 |
a and c got to be 1 and 2 since some of all would not exceed 298 and given the sum conditions
16 + 63 + 29 + 95 = 203
29 + 63 + 13 = 105
i have not used 8 in my solution
x and w are not written in down -
Anil Jain- | a | b |
c | - | - |
d | - | - |
Given that a, c , d, c, can not be 0;
For the rules its aparent that a, b , c, d are two digits ages only,
Now, C has to be less than 399+; again, C is sum of all four ages and a is sum of 3 ages, so a is less than c
Hence C has to be 2 and a has to be 1
Now as c = W+X+Y+Z which is a three digit figure, so ony following combination seems possible which will satisfy all the conditions****
- | 1 | 6 |
2 | 0 | 0 |
9 | 5 | - |
Hence:
w=16
x=95
y=60
z=29
c (Across)=w+x+Y+z = 200
a (Down)=W=Y+Z = 105
Hence w and x are written across.
-CB -
silverscorpion|-|1|6|
|2|0|4|
|9|7|-|
The middle number is common between C-across and A-down. This means that the difference between the sum of 4 ages and that of 3 ages is in the nineties.
The maximum it can be is 98, (since digits should not be repeated).
But whatever the case, we can see that D-across is where that left out age will come.
Nevertheless, how much ever I try, I can't seem to accurately fill the boxes. There is a difference of 2 between the actual sum of 4 ages and C-across in the previous box.
i.e., 29 + 97 + 16 + 64 = 206, whereas C-across is 204.
This is the closest approx. i could come up with.
So, to answer the question, X's age was left out in A-down. -
Ankita KatdareThis question has been declared as invalid.
I apologize to all the participants.
This is one of my biggest goof-ups till date.
Coins of this question will not be credited to anyone.
I feel ashamed of my activity. ๐
PS: I again say sorry for behaving in a crude manner is preparing this question. ๐ญโ
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