Question

Let $m$ be an integer such that $1 \leq m \leq 1000$. The probability of selecting at random an integer $m$ such that the quadratic equation $6 x^2-5 m x+m^2=0$ has at least one integral solutions, is

• A. $\frac{333}{1000}$
• B. $\frac{500}{1000}$
• C. $\frac{833}{1000}$
• D. $\frac{667}{1000}$

Answer 1

Answer: (D)

The quadratic equation has two solutions
$x =\frac{ m }{2} ; \quad x =\frac{ m }{3} \quad \text { (using } x =\frac{- b \pm \sqrt{ b ^2-4 a c}}{2 a } \text { ) }$
There are 500 multiples of 2, 333 multiples of 3 , and 166 multiples of 6 between 1 and 1000 . Therefore, the probability is
$ p =\frac{500+333-166}{1000}=\frac{667}{1000} $