Arrange the numbers
1
−
1
0
without repetition to replace the alphabets in the grid such that the sum of the numbers in the top row, bottom row, left column and right column is
1
8
each.
Among the solutions that exist, the alphabets E and F can each have only one value, interchangeable between them.
Find E × F
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A + B + C + D + E + F + G + H + I + J = 5 5
A + B + C + D + G + H + I + J = 3 6
⇒
E + F = 1 9 and 2 1 9 = 9 . 5 ⇒ E = 1 0 , 9 and F = 9 , 1 0
E × F = 9 × 1 0 = 9 0
Since the sum of the numbers in the left column and the sum of the numbers in the right column are both 18, then A + B + C + D + E + G + H = 3 6 .
Using the fact that 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 0 = 5 5 , we will get E + F = 1 9 .
The only possible value for E and F is 9 and 10.
Therefore E × F = 9 0
The horizontal rows on the top and bottom have 1 less space, so I placed the largest numbers in there. You don't want the 2 largest numbers to be in the corners because then one column would be fulfilled, but then the other column wouldn't have enough numbers to fill up. Without filling up the whole grid, you can deduce that the answer is 10X9=90.
I did it in a bit different way. The circled term in the figure are the two greatest numbers, so its pointless to keep them in corners, second point is that it is also pointless to keep them in the two columns as if we fit them there then one column will get complete, but other won't satisfy. Now first I completed the upper and lower rows as shown by (2,6) and (8,1) now the remaining four numbers of the columns i.e. (5,3,4,7) can be fitted very easily
.
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The only other solution will have 2 , 6 and 5 , 4 making up the corners.