markr wrote:DOMINOES
12, 22, 44, 34
Code: ---------
|1|1 2|2 2|
--- ---
|4|4 4|3 4|
- --- ---
|3|1|2|1|2|
|3|1|4|3|3|
---------
Just brilliant! How could he come up with the answer so quick? Well, one thing fer sure; he didn't do it this way:
First of all, take a look at what the 10 dominoes look like:
1 1 -- 2 2 -- 3 3 -- 4 4 -- 1 2 -- 1 3 -- 1 4 -- 2 3 -- 2 4 -- 3 4
Notice from this that each number appears 5 times, as this will become important further on.
Now we take a look at the possible ways of reaching each required number (the numbers that each row or column must equal when multiplied)
8
1 x 1 x 1 x 2 x 4
2 x 2 x 2 x 1 x 1
768
4 x 4 x 4 x 4 x 3
12
1 x 1 x 1 x 4 x 3
1 x 1 x 3 x 2 x 2
108
3 x 3 x 3 x 4 x 1
3 x 3 x 3 x 2 x 2
36
4 x 3 x 3 x 1
3 x 3 x 2 x 2
4
1 x 1 x 1 x 4
1 x 1 x 2 x 2
64
4 x 4 x 4 x 1
2 x 2 x 4 x 4
18
3 x 3 x 2 x 1
48
4 x 4 x 3 x 1
4 x 3 x 2 x 2
Next up is to try and look at lines going across and down to see if there are digits that can only go in a certain place in order to satisfy both the across and down lines of numbers. The first lines I noticed were the 4 and 768 lines.
The 768 line only contains the digits 3 and 4, and the 4 line contains only 1 and 4 or only 1 and 2. So we can rule out the 1 x 1 x 2 x 2 = 4 line since it doesn't contain a 3 or 4. And we also know the position of the 4 in the 1 x 1 x 1 x 4 line. Since this line only contains one 4, it must go in the 768 line, because the 768 line does not contain a 1. And we can then also fill in all the other digits in the 4 line as they are all 1's.
So this is our grid so far:
? 1 ? ? ?
? 4 ? ? ?
? 1 ? ? ?
? 1 ? ? ?
Next up is to look at the 108 line. We know from the grid it must contain a 1, and so we know that our 108 line is a variation of 3 x 3 x 3 x 4 x 1
Next up is the 64 line. None of the two possible combinations contain a 3, and so we know that the 3rd digit of the 768 line must be a 4. Next up is the 18 line which is 3 x 3 x 2 x 1. Since the 768 line does not contain a 2 or 1, we know that a 3 must be the 4th digit of the 768 line. And so from that we can fill in the whole of the 768 line. So this is now our grid:
? 1 ? ? ?
4 4 4 3 4
? 1 ? ? ?
? 1 ? ? ?
Now we look at the 36 line. We know it must contain a 4, so we know its line is a combination of 4 x 3 x 3 x 1. Now we look at the 8 line. None of the possible answers contain a 3. Therefore, we know that the first digit od the 8 line must be a 1. We know this because it's he only number possible that will also satisfy the 36 line vertically. And we can also fill in the rest of the 36 line now. So here's our grid again:
1 1 ? ? ?
4 4 4 3 4
3 1 ? ? ?
3 1 ? ? ?
Next we look at the 18 line and the 108 line. The rest of the digits for the 18 line are 3, 2, 1 and the rest of the digits for the 108 line are 4, 3, 3. From this we can see that a 3 must go as the 4th number in the 18 line. Here's the grid now:
1 1 ? ? ?
4 4 4 3 4
3 1 ? ? ?
3 1 ? 3 ?
Next we look at the 108 and 64 lines. The rest of the digits for 108 are 4, 3 and the rest for 64 are 4, 2, 2. The only common number here is 4, and so this must go as the 3rd digit of the 108 line. And so we can also fill in the rest of the 108 line now. So here's our grid now:
1 1 ? ? ?
4 4 4 3 4
3 1 ? ? ?
3 1 4 3 3
Now is the time to look at how many times each number has appeared, remembering that each must appear 5 times. So far we have:
1 - 4 times
2 - 0 times
3 - 5 times
4 - 5 times
We can now look at all the possible combinations we have left still and cross off any that contain another 3 or 4. So we can get rid of the 1 x 1 x 1 x 2 x 4 = 8 line. We know the 8 line is 2 x 2 x 2 x 1 x 1. Since we have the first two digits already filled in, we know the rest of the 8 line are all 2's. And the 64 line now appears as 2 4 ? 4, so we know that the missing digit is 2 for this. And our 48 line at the moment is 2 4 ? 3 so we know the missing digit is a 2. This leaves 1 box of the grid left to be filled in. We know it must be a 1 because we still need to have another 1 digit to make it appear 5 times. So here's our final grid layout:
1 1 2 2 2
4 4 4 3 4
3 1 2 1 2
3 1 4 3 3
Now it's just a case of positioning the dominoes correctly, remembering that only 4 are placed horizontally. I think this is just trial and error getting the right positions.
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