2015 CSMC Problem PartA #6

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 $12$ , since there are $6$ pairs and $2$ values ( $\tan X$ and $\tan Y$ ), so $2 \times 6 = 12$ $\\$ //===== $\\$ Legit Solution Starts Here $.\\$ $\tan (X+Y) = \tan 45º$ $\\$ $\tan(X+Y) = 1$ $\\$ $\frac {\tan X + \tan Y}{1-\tan X + \tan Y} = 1$ $\\$ $1 = \frac {\frac 1m + \frac an}{1 - \frac {a}{mn}}$ $\\$ Simplify and get: $\\$ $0 = am + a+n-nm$ Rearrange and get $a+a = m(n-a)-(n-a)$ $\\$ $2a = (m-1)(n-a)$ $\because m \neq 1, m-1 > 0, 2a > 0 \quad \therefore n>a$ $\\$ $.\\$ Using variable substitution $\\$ Let $m - 1 = r > 0, n-a = s > 0$ $\\$ $2a = r \cdot s$ $\\$ $\because 2a$ has $3$ factorizations $(r, s, rs)$ $\therefore$ it has $3! = 6$ positive divisors $\\$ $.\\$ By Fudemental Theorem of Arithmetic $\\$ Given integer $d$ and its prime factorizations $d = p_1^{c_1} \times p_2^{c_2} \times p_3^{c_3} \times ... \times p_k^{c_k}$ 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 $(c+1) = 6 \Rightarrow c = 5, d = p^{c} = p^5$ ] $\\$ or the form $p_1^2 \cdot p_2$ for some $prime numbers$ $p_1$ and $p_2$ $\\$ $.\\$ By Property of Composite Numbers $\\$ If $2a$ is of the form $p^5$ , then $p = 2$ since $2a$ is already divisible by 2 and we can only pick one prime number. $\\$ Thus $2a = 2^5 = 32$ , so $a = 16$ $\\$ $.\\$ If $2a$ is of the form $p_1^2 \cdot p_2$ , then $p = 2$ or $q = 2$ [as $2$ must be one factorization for $2a$ ]. $\\$ Since $a < 50$ , $\\$ If $p_1 = 2$ , then $2a = 4 \cdot p_2$ . Since $a \leq 50$ , then $2a \leq 100$ or $p_2 \leq 25$ , and so $p_2$ can equal $3, 5, 7, 11, 13, 17, 19, 23$ giving $a = 6, 10, 14, 22, 26, 34, 38, 48 (p_2 \neq 2)$ $\\$ If $p_2 = 2$ , then $2a = 2p_1^2$ since $2a \leq 100$ , then $p_1^2 \leq 50$ , and so the prime $p_1$ can equal to $3, 5, 7$ giving $a = 9, 25, 49 (p_1 \neq 2)$ . $\\$ $.\\$ There fore, in summary, there are $\boxed{12}$ possible $a$ .

Base on the official solution by University of Waterloo $\\$ All Credit Goes to University of Waterloo