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124C41
 Senior Boarder
Posts: 63 
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Enigma 1314 - Times table New Scientist magazine, 6 November 2004. by Susan Denham.
From a full set of dominoes, I have taken just those that have a 1, 2, 3 or 4 at each end. I then arranged them into a rectangle, as shown,
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Soultra
Expert Boarder
Posts: 91
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Peter Chapman quotes Enigma 1314 by Susan Denham:
Answer: 1-2, 2-2, 3-4, and 4-4.
Full layout after 36 lines: 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2
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glundby
Expert Boarder
Posts: 93
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(I'm numbering the columns 1-5 from left to right, and the rows 1-4 from top to bottom.)
A quick glance at the 10 dominoes shows there's five 1's, five 2's, five 3's, and five 4's.
We see from the products that there must be two 3's in the first column, two in the fourth, and one in the fifth.
There are three 3s in the fourth row, 1 in the third and 1 in the second.
The only combination that works for the second row is four 4's and a 3. Therefore, the second and third squares in row two are 4's. There are then 1's in the other three squares of column 2.
This means that the only combination of numbers for row 4 is three 3's, a 4, and a 1, and we can fill them in in the appropriate places:
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iphwin
Expert Boarder
Posts: 83
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'Trial and error' suggests an unfocused process, but of course the idea is to find places where there are only a few choices to consider. This was my path to the solution; there may very well be others of similar length. Having reached the layout of 20 numbers (by the reasoning that Ted details
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MishaEE
Senior Boarder
Posts: 68
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The dominoes are 11 12 13 14 22 23 24 33 34 44.
Label the rows 1 to 4 from top to bottom. Label the columns 1 to 5 from left to right.
Consider the rows: 8 = 2^3 -> 22211 or 42111 768 = 2^8 * 3 -> 34444 12 = 2^2 * 3 -> 32211 or 43111 108 = 2^2 * 3^3 -> 33322 or 43331
Consider the columns: 36 = 2^2 * 3^2 -> 3322 or 4331 4 = 2^2 -> 4111 or 2211 64 = 2^6 -> 4441 or 4422 18 = 2 * 3^2 -> 3321 48 = 2^4 * 3 -> 4322 or 4431
The four 4's in row 2 must go in columns 1, 2, 3, 5. The three 3's in row 4 must go in columns 1, 4, 5. Column 1 must be 4331. Column 2 must be 4111. The 3 in column 1 must go in row 3. Row 4 must be 43331. We have used up all five 4's. Row 1 must be 22211. Row 3 must be 32211. Column 3 must be 4422. Column 5 must be 4322. Row 3's one unplaced 1 must go in column 4. All five unfilled squares must be 2's.
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