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# I need help on this problem.

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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.

Oct 12, 2018

#2
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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!

Oct 12, 2018

#1
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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%

Oct 12, 2018
edited by Guest  Oct 12, 2018
#2
+269
+2

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!

ANotSmartPerson Oct 12, 2018