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Answers
Given : 9 Empty boxes ( 3 * 3) & their multiplication output across row & columns
To find : Fill the boxes with 1 to 9 Digits without any repetition so that all rows & column satisfied
Solution:
5 & 7 are prime numbers whose multiples are not available from 1 to 9
Let see which numbers should have these factors
21 & 56 should have 7 as Factor
Hence 3rd row last column must be 7 as that is common factor of both
120 * 180 should have 5 as Factor
Hence 2nd row 2nd column must have 5 as that is common factor of both
21 = 7 * 1 * 3 ( only possible)
but if we use 1 in 2nd row , 3rd column then
24 * 1 * 5 = 120 ( not possible)
Hence
3 will be in 2nd row , 3rd column & 1 will be in 1st row , 3rd column
2nd row 1st column = 120 / (3 * 5) = 8
Solving this way we can reach that
6 X 9 X 1 = 54
X X X
8 X 5 X 3 = 120
X X X
2 X 4 X 7 = 56
= = =
96 180 21
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