Using all the numbers {1,2,3,4,5,6,7,8}, each exactly once, fill each empty white square with a number so that all the horizontal and vertical equations are true. The grey squares are not used.
Attribution: Erich Friedman
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asked Sep 3 at 6:42
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3 Answers
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The filled grid:
Explanation:
The 1 cannot be part of a multiplication or division without causing duplicate numbers, and cannot be the result of an addition. The only remaining option is at the start of the second row.
Only 2 can work as a divisor, because anything larger will make the division equal 2 and then cause problems with the multiplication in the second column becoming too large.
The 8 now must go in the bottom of the second column, as anywhere else rapidly leads to numbers becoming too large.
The rest then follows easily.
answered Sep 3 at 7:00
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1
The first row has very few possibilities: 8,2,4; 8,4,2; 6,2,3; 6,3,2. But 8 as the dividend doesn't allow another addend in the first column, so it's 6,3,2 or 6,2,3. The other addend in the first column can't be 2 or 3, then, and 4 is too large, so it's 1. That means that the other multiplicand in the second column is not 1, 2, or 3, so it must be 4 to keep the product small enough: and then the first multiplicand is 2. So we get
6 2 3
1 4 5
7 8
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3
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Consider
the middle number in the first row. It equals the ratio between the other numbers in the first row, and also between the other numbers in the second column. 1 is inadmissible, because the numbers are all distinct, and every number larger than 2 has at most one pair of numbers greater than one with that ratio, so 2 goes in that spot.
3 and 4 now occupy the top right and central spaces, and 6 and 8 occupy the top left and bottom middle spaces, leaving 1, 5, and 7 to be placed. These must go in the center left, center right, and bottom left spaces, respectively, and the rest follows.
6 2 3 1 4 5 7 8
