How do you approach problems that are seemingly ridiculous?

<p>Too lazy to think about the first one (god i hate probability&counting), but here’s the answer to the second.</p>

<p>Foil the first part to get:
x^2 +bx + ax + ab = x^2 +cx + d
Therefore, bx+ax=cx and ab=d
Simplify to get b+a=c and ab=d</p>

<p>Because d is a prime number, a or b has to be 1 while the other number must equal d. Since b is greater than a, a=1 and b=d. Choice A is out. b/d must equal a by ab=d, and a=1. Choice b is out. Since b=d, you can substitute d for b in the first equation, which yields d+a=c. Thus, c-d=a, and a=1. Choice C is out. Divide both sides of the first equation by (a+b). Choice D is out. E is all that remains.</p>

<p>Condensing Latency’s solution.</p>

<p>Without distributing everything you can tell ab=d.

Put that in (E): d-b=0, so d-b=/=1.</p>

<p>@smash20
Your solution for (1.) is correct (provided you quoted the question correctly):
2 can not eat the driver’s eat. :P</p>

<p>And btw - why bother with PR?</p>

<p>Where are all these questions from?</p>