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Question 83
The number of natural numbers k such that $$\frac{3k^2 + 4k + 12}{k}$$ is a prime is
Solution
For the result to be prime, the expression must firstly yield a whole number.
The expression can be broken up as = 3*k + 4 + (12/k)
Hence, k has to be a whole number factor of 12.
Probable values of k = 1,2,3,4,6,12
Of these, the values of k = 1,3,4,12 yield prime numbers for the expression.
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