Question

Constantine forms the following hypothesis. Let n be any non-negative number that meets the following condition: when n is divided by 5, the remainder cannot equal 2. For such values of n, the quantity Q = 97 − 6n is a prime number so long as Q > 0. Which of the following values of n would provide a counterexample to this hypothesis? Indicate all such values.

Answer 1

To find a counterexample, one must determine a value of n satisfying the condition of the remainder not being 2 when divided by 5, but resulting in a composite number for Q = 97 - 6n. Checking n=3, n=10, and n=15 results in prime numbers for Q, so one must continue the process until a composite Q is found to disprove the hypothesis.

Step-by-step explanation:

The student's question asks for a counterexample to disprove Constantine's hypothesis, which states that for non-negative numbers n that when divided by 5 give a remainder not equal to 2, the expression Q = 97 − 6n will result in a prime number, provided that Q > 0. To find such a counterexample, one must identify a value of n that meets the condition but results in Q being a composite number.

Upon checking various values that satisfy the condition, one may eventually find that for n=3 (which when divided by 5, gives a remainder of 3, thus meeting the condition), Q equates to 97 − (6 × 3) = 97 − 18 = 79, which is prime, so it does not serve as a counterexample. However, for n=10 (which also meets the condition), Q equates to 97 − (6 × 10) = 97 − 60 = 37, which is also prime. Continuing this process and checking n=15, results in Q equating to 97 − (6 × 15) = 97 − 90 = 7, again, a prime number. Eventually, if the student finds a value of n that results in Q being a composite number, it will serve as the counterexample needed to disprove the hypothesis.