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There are 3 datatables. We forgot the numbers around the third square. How can we compute the 6 missing numbers?(there are question marks instead of them)

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  • 2
    $\begingroup$ Are those things on the border of square 3 that look like 2's the question marks? This belongs on puzzling.se $\endgroup$ – Ross Millikan Dec 2 '14 at 17:19
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    $\begingroup$ 142 down the left, 677 along the bottom. This is a way of multiplying by hand. $\endgroup$ – Barry Cipra Dec 2 '14 at 17:23
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    $\begingroup$ @BarryCipra A lovely, but damn old way :-) Veritas, have also a look at lattice multiplication $\endgroup$ – Jean-Claude Arbaut Dec 2 '14 at 17:25
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    $\begingroup$ @Joe I'm curious, is asking for an explanation of the answer to a mathematical puzzle not on-topic here? $\endgroup$ – Aza Dec 2 '14 at 18:28
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    $\begingroup$ This question appears to be off-topic because it is about long multiplication. Basic well-known math systems are not puzzles. 2+2=? is not a puzzle, nor is long multiplication $\endgroup$ – Joe Dec 2 '14 at 18:34
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Moving top to bottom on the left side...

1 4 2

Going right to left on the bottom...

6 7 7

These values in the third table, using the same method as the other two tables...

...multiply two three-digit values. So the third one shows that 249 times 573 equals 142,677 (value on left, read downward, concatenated with value on bottom, read going right). The technique is called "lattice multiplication." It works because it's simply like multiplying any long numbers by hand. That is, multiplying each combination of digits and summing for the appropriate values of the answer's digits in their corresponding places. It's just a different format of writing out the calculation.

The values are calculated as follows:

The value in each cell is its column's (top) header value multiplied by the header value to the right of its row. Then, the product's two digits are split by a diagonal line which bisects the cell equally from top-right to bottom-left. After that, the values in each diagonal stripe, starting on the bottom-right and moving to the top-left, are summed up, carrying any overflow into a second digit into the sum of the next stripe. Those summed values are placed either to the bottom of the table or to its left, depending on where the diagonal stripe for the summed values end.

[EDIT:]

Thought of an easier solution:

Just multiply the number on top of the table, read moving to the right, by the number to the right, read going down. Then, the first three digits of the product go top to bottom on the left, and the next three digits go left to right on the bottom. (The question is how the numbers can be computed - not that they have to be computed in the way implied by the table's presence, right?)

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  • $\begingroup$ If it does stay here, it needs an answer posted as such, yeah? $\endgroup$ – tjbtech Dec 2 '14 at 18:42
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These tables are examples of lattice multiplication.

Ignoring the diagonal lines, the numbers in each of the squares is the product of the number at the top of its column and the number at the right of its row. For example, in the top-left of the first example in the question, $4 \cdot 8 = 32$.

          4
+---+---+---+
|   |   |   |
|   |   |32 | 8
|   |   |   |
+---+---+---+
|   |   |   |
|   |   |   |
|   |   |   |
+---+---+---+
|   |   |   |
|   |   |   |
|   |   |   |
+---+---+---+

Then, in the square, write the tens digit of the product to the left of the diagonal line and the ones digit to the right of it.

          4
+---+---+---+
|  /|  /|3 /|
| / | / | / | 8
|/  |/  |/ 2|
+---+---+---+
|  /|  /|  /|
| / | / | / |
|/  |/  |/  |
+---+---+---+
|  /|  /|  /|
| / | / | / |
|/  |/  |/  |
+---+---+---+

Then, looking at the diagonal lines and ignoring the vertical and horizontal lines making the squares, we see that there are several diagonal columns going down and to the left.

  2   7   4
+---+---+---+
|1 / 1 / 3 /|
| /   /   / | 8
|/ 6 / 4 / 2|
+   /   /   +
|1 / 6 / 3 /|
| /   /   / | 9
|/ 8 / 3 / 6|
+   /   /   +
|1 / 3 / 2 /|
| /   /   / | 5
|/ 0 / 5 / 0|
+---+---+---+

For each of these diagonal columns, starting with the leftmost, add up all of the digits that are in them. Write the ones digit of the sum in the space immediately to the bottom or left of the column (where the question marks are in example 3), then carry the tens and larger digits to the next column.

In example three, the multiplication steps have already been completed. All that's left is to add up the diagonal columns. Doing so, we find that the numbers in the question marks are:

    2   4   9
  +---+---+---+
* |1 /|2 /|4 /|
1 | / | / | / | 5
* |/ 0|/ 0|/ 5|
  +---+---+---+
* |1 /|2 /|6 /|
4 | / | / | / | 7
* |/ 4|/ 8|/ 3|
  +---+---+---+
* |0 /|1 /|2 /|
2 | / | / | / | 3
* |/ 6|/ 2|/ 7|
  +---+---+---+
   *6* *7* *7*

Since lattice multiplication is an algorithm for multiplying, we can find the numbers in the question marks even faster: Ignore the numbers in the lattice altogether, multiply the numbers going across the top and right of the lattice, and write the product along the left and bottom of the lattice. In the case of example three, the number going across the top of the lattice is $249$, the number going down the right of the lattice is $573$, and their product is $249 \cdot 573 = 142677$.

    2   4   9
  +---+---+---+
* |   |   |   |
1 |   |   |   | 5
* |   |   |   |
  +---+---+---+
* |   |   |   |
4 |   |   |   | 7
* |   |   |   |
  +---+---+---+
* |   |   |   |
2 |   |   |   | 3
* |   |   |   |
  +---+---+---+
   *6* *7* *7*
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