Question
How many pairs of positive integers (a, b) are there such that a and b have no common factor greater than 1 and ab+14b9a is an integer?
Answer: Option D
:
D
Let u=ab. Then the problem is equivalent to finding all possible rational numbers 'u' such that
u+149u=K, for some integer K.
This equation is equivalent to 9u2−9uk+14=0
9u2−9uk+14=0 whose solutions are,
u=9k±√81k2−50418
=K2±16√9k2−56
Hence u is rational if and onlly if
√9k2−56 is rational, Which is true if and only if 9k2−56 is a perfect square. Suppose that 9k2−56=S2 for some positive integer S.
(3k+s)(3k-S)=56
The only factors of 56 are 1,2,4,7,8,14,28 and 56. So, (3k-S) and (3k+s) is one of the
ordered pairs (1,56), (2,28), (4,14) and (7,8). The cases (1,56) and (7,8) yield no (4,14) yield k=5 and k=3 respectively.
If k=5 then u=13 or u=143
If k=3 then u=23 of u=73.
Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3)
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:
D
Let u=ab. Then the problem is equivalent to finding all possible rational numbers 'u' such that
u+149u=K, for some integer K.
This equation is equivalent to 9u2−9uk+14=0
9u2−9uk+14=0 whose solutions are,
u=9k±√81k2−50418
=K2±16√9k2−56
Hence u is rational if and onlly if
√9k2−56 is rational, Which is true if and only if 9k2−56 is a perfect square. Suppose that 9k2−56=S2 for some positive integer S.
(3k+s)(3k-S)=56
The only factors of 56 are 1,2,4,7,8,14,28 and 56. So, (3k-S) and (3k+s) is one of the
ordered pairs (1,56), (2,28), (4,14) and (7,8). The cases (1,56) and (7,8) yield no (4,14) yield k=5 and k=3 respectively.
If k=5 then u=13 or u=143
If k=3 then u=23 of u=73.
Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3)
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