Answer:
The statement 2 alone is sufficient.
Explanation:
Using the Arithmetic Properties of numbers we have
Given that x is a multiple of 6, x has at least 1 factor of 2 and at least 1 factor of 3. Given that y is a multiple of 14, y has at least 1 factor of 2 and at least 1 factor of 7. So, xy has at least 1 factor of 2, at least 1 factor of 3, and at least 1 factor of 7 and can be expressed as xy = (2)(3)(7)r = 42r, where r is a positive integer. Determine if xy is a multiple of 105 = (3)(5)(7).
1) If x = (2)(3)(3), then x is a multiple of both 6 and 9. If y = (2)(7) then y is a multiple of 14. In this case, xy = (2)(3)(3)(2)(7) = (2²)(3²)(7) and is not a multiple of 105. However, if x = (2)(3)(3)(5), then x is a multiple of both 6 and 9. If y = (2)(7), then y is a multiple of 14. In this case, xy =(2)(3)(3)(5)(2)(7) = (2²)(3²)(5)(7) and is a multiple of 105; NOT SUFFICIENT.
2) If y = (2)(5)(5)(7), then y is a multiple of 14 and 25. If x is a multiple of 6, x = (2)(3)q for some positive integer q. Then xy = (2)(3)q(2)(5)(5)(7) = (2)(2)(3)(5)(5)(7)q and xy is a multiple of 105; SUFFICIENT.
The statement 2 alone is sufficient.