For natural numbers x, y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is

Options

1. 34
2. 50
3. 42
4. none

Correct Answer

34

Explanation

y(x+z)=19 and z(x+y)=51. Since 19 is prime and x+z≥2, we must have y=1 and x+z=19. With y=1, z(x+1)=51. But x=19−z, so z( (19−z) + 1 ) = 51 ⇒ z(20−z)=51 ⇒ z^2 −20z +51 = 0. Solve: Δ = 400 − 204 = 196, √Δ = 14. z = (20 ± 14)/2 ⇒ z = 17 or z = 3. If z=17 then x = 19−17 = 2 and xyz = 2·1·17 = 34. If z=3 then x = 19−3 = 16 and xyz = 16·1·3 = 48. Minimum possible value of xyz = 34.

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