Suppose that and are angles with and for some positive integers , and . Determine the number of positive integers for which there are exactly pairs of positive integers with .
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This question seems to be a trigonometry question but it is actually a number theory question //===== If I am to take a theoretical guess the answer would be 1 2 , since there are 6 pairs and 2 values ( tan X and tan Y ), so 2 × 6 = 1 2 //===== Legit Solution Starts Here . tan ( X + Y ) = tan 4 5 º tan ( X + Y ) = 1 1 − tan X + tan Y tan X + tan Y = 1 1 = 1 − m n a m 1 + n a Simplify and get: 0 = a m + a + n − n m Rearrange and get a + a = m ( n − a ) − ( n − a ) 2 a = ( m − 1 ) ( n − a ) ∵ m = 1 , m − 1 > 0 , 2 a > 0 ∴ n > a . Using variable substitution Let m − 1 = r > 0 , n − a = s > 0 2 a = r ⋅ s ∵ 2 a has 3 factorizations ( r , s , r s ) ∴ it has 3 ! = 6 positive divisors . By Fudemental Theorem of Arithmetic Given integer d and its prime factorizations d = p 1 c 1 × p 2 c 2 × p 3 c 3 × . . . × p k c k where p 1 , p 2 , . . . p k are distinct prime number and c 1 , c 2 , . . . c k are positive integers. . By Basic Counting Technique The number of positive divisor of d is ( c 1 + 1 ) ( c 2 + 1 ) . . . ( c k + 1 ) , which is 6 In order for this product to equal to 6 , d must have the form p 5 [Facorizations are ( c + 1 ) = 6 ⇒ c = 5 , d = p c = p 5 ] or the form p 1 2 ⋅ p 2 for some p r i m e n u m b e r s p 1 and p 2 . By Property of Composite Numbers If 2 a is of the form p 5 , then p = 2 since 2 a is already divisible by 2 and we can only pick one prime number. Thus 2 a = 2 5 = 3 2 , so a = 1 6 . If 2 a is of the form p 1 2 ⋅ p 2 , then p = 2 or q = 2 [as 2 must be one factorization for 2 a ]. Since a < 5 0 , If p 1 = 2 , then 2 a = 4 ⋅ p 2 . Since a ≤ 5 0 , then 2 a ≤ 1 0 0 or p 2 ≤ 2 5 , and so p 2 can equal 3 , 5 , 7 , 1 1 , 1 3 , 1 7 , 1 9 , 2 3 giving a = 6 , 1 0 , 1 4 , 2 2 , 2 6 , 3 4 , 3 8 , 4 8 ( p 2 = 2 ) If p 2 = 2 , then 2 a = 2 p 1 2 since 2 a ≤ 1 0 0 , then p 1 2 ≤ 5 0 , and so the prime p 1 can equal to 3 , 5 , 7 giving a = 9 , 2 5 , 4 9 ( p 1 = 2 ) . . There fore, in summary, there are 1 2 possible a .
Base on the official solution by University of Waterloo All Credit Goes to University of Waterloo