Suppose that 4 x 1 = 5 , 5 x 2 = 6 , 6 x 3 = 7 . . . 1 2 6 x 1 2 3 = 1 2 7 , 1 2 7 x 1 2 4 = 1 2 8 . If the product x 1 x 2 x 3 . . . x 1 2 3 x 1 2 4 is of the form n m , g cd ( m , n ) = 1 , m ∈ Z + , n ∈ Z + , find m + n .
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I was trying to take the base change log formula, but this is much more elegant!
4 x 1 = 5 ⟹ x 1 = lo g 4 5 , 5 x 2 = 6 ⟹ x 2 = lo g 5 6 ⋯ 1 2 7 x 1 2 4 = 1 2 8 ⟹ x 1 2 4 = lo g 1 2 7 1 2 8
Product = x 1 × x 2 × ⋯ × x 1 2 4
Product = lo g 4 5 × lo g 5 6 × ⋯ × lo g 1 2 7 1 2 8
Product = lo g ( 4 ) lo g ( 5 ) × lo g ( 5 ) lo g ( 6 ) × ⋯ × lo g ( 1 2 7 ) lo g ( 1 2 8 )
Product = lo g ( 4 ) lo g ( 5 ) × lo g ( 5 ) lo g ( 6 ) × ⋯ × lo g ( 1 2 7 ) lo g ( 1 2 8 )
Product = lo g ( 4 ) lo g ( 1 2 8 ) = lo g 4 1 2 8
Product = 2 7 ⟹ m = 7 , n = 2
⟹ m + n = 7 + 2 = 9
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1 2 7 x 1 2 4 = 1 2 8 ( 1 2 6 x 1 2 3 ) x 1 2 4 = 1 2 6 x 1 2 3 x 1 2 4 = 1 2 8 ( 1 2 5 x 1 2 2 ) x 1 2 3 x 1 2 4 = 1 2 8 Doing this many times you get 4 x 1 x 2 x 3 x 4 . . . x 1 2 3 x 1 2 4 = 1 2 8 ( 2 2 ) x 1 x 2 x 3 x 4 . . . x 1 2 3 x 1 2 4 = 2 2 ( x 1 x 2 x 3 x 4 . . . x 1 2 3 x 1 2 4 ) = 2 7 ⇒ 2 x 1 x 2 x 3 x 4 . . . x 1 2 3 x 1 2 4 = 7 x 1 x 2 x 3 x 4 . . . x 1 2 3 x 1 2 4 = 2 7 g cd ( 7 , 2 ) = 1 ; 2 ∈ Z ; 7 ∈ Z 7 + 2 = 9