Bunuel
A prime number n is a factor of both 14k + 13 and 7k + 1, where k is a positive integer. What is the value of n?
(A) 2
(B) 3
(C) 7
(D) 11
(E) 13
Questions dealing with number properties, factorizations and remainders can often be solved with logic and very few calculations.
We'll look for such a solution, a Logical approach.
Since n is a factor of 7k+1 it must also be a factor of 2*(7k+1) = 14k+7.
Since n is a factor of both 14k+13 and 14k+7 it is also a factor of the difference between them, that is of (14k+13)-(14k+7) = 6.
Then n can be 1,2,3,6 but since it is prime it can only be 2 or 3.
To decide between 2 and 3 we'll need to see if have any even/odd properties.
14k is even and 13 is odd so 14k+13 must be odd. That is, n is a factor of an odd number and cannot be 2.
(B) is our answer.
Isn't 2(7k+1) = 14k + 2? And therefore difference will be 11. I got 11 by substituting possible values for k.
Please correct me if I am wrong.