+283 I randomly pick an integer between 1 and 10 inclusive. What is the probablility that I choose a p such that there exists an integer q so that p and q satisfy the equation pq-4p-2q=2? Express your answer as a common fraction.
+283 Thank you for trying, but the answer was incorrect :( Here is the solution I got back in case it helps.
We approach this problem by trying to find solutions to the equation pq-4p-2q=2 . To do this, we can use Simon's Favorite Factoring Trick and add 8 to both sides to get qp-4p-2q+8=10. This can be factored into
(p-2)(q-4)=10
We now can see that there are solutions only if p-2 divides 10 . Thus, there are 4 possible values of p between 1 and 10 inclusive (1,3,4,7) . It follows that the probability of picking such a p is 2/5.
Thanks again!
There are 3 numbers for p and corresponding 3 numbers for q that will satisfy your equation as follows:
p = (2 (q + 1))/(q - 4) and q must not =4
q = (4 p + 2)/(p - 2) and p must not=2
p =3 and q=14
p =4 and q=9
p =7 and q=6
Therefore the probability is:
3/10 =30%
+283 Thank you for trying, but the answer was incorrect :( Here is the solution I got back in case it helps.
We approach this problem by trying to find solutions to the equation pq-4p-2q=2 . To do this, we can use Simon's Favorite Factoring Trick and add 8 to both sides to get qp-4p-2q+8=10. This can be factored into
(p-2)(q-4)=10
We now can see that there are solutions only if p-2 divides 10 . Thus, there are 4 possible values of p between 1 and 10 inclusive (1,3,4,7) . It follows that the probability of picking such a p is 2/5.
Thanks again!
