For positive rational , how many ordered pairs of positive integers are possible?
This problem is not original
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lo g a x + lo g a y lo g a ( x ⋅ y ) x y 2 0 2 0 Factors of 2020 = lo g a 2 0 2 0 = lo g a 2 0 2 0 = 2 0 2 0 = 2 2 ⋅ 5 1 ⋅ 1 0 1 1 = ( 2 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 1 2
Since each factor corresponds to another factor to multiply to 2020, there are 12 pairs in total
1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 1 0 . 1 1 . 1 2 . x 1 2 4 5 1 0 2 0 1 0 1 2 0 2 4 0 4 5 0 5 1 0 1 0 2 0 2 0 y 2 0 2 0 1 0 1 0 5 0 5 4 0 4 2 0 2 1 0 1 2 0 1 0 5 4 2 1