Science:Math Exam Resources/Courses/MATH312/December 2012/Question 02 (a)

MATH312 December 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q3  •  Q4  •  Q5  •  

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Question 02 (a)
State whether the following are true or false with full justification. Each question carries 4 marks. If a positive integer has exactly 3 positive divisors, then it is necessarily of the form $p^{2}$ where p is a prime.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Suppose a number has at least two distinct prime divisors. What is a lower bound on the number of divisors in this case?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The answer is true. Suppose that n has at least two distinct prime divisors p and q. Then it has at least 4 positive divisors given by $\displaystyle 1,p,q,pq$. Thus n must only have one prime divisor, so must be of the form $\displaystyle n=p^{\alpha }$. In this case, n has exactly $\displaystyle \alpha +1$ positive divisors, namely, $\displaystyle p^{\beta }$ for $\displaystyle \beta =0,1,2,\ldots ,\alpha$. Thus if $\displaystyle \alpha +1=3$ we have that $\displaystyle \alpha =2$ which is as claimed in the question.

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