Science:Math Exam Resources/Courses/MATH312/December 2009/Question 01 (a)

MATH312 December 2009

  •  Q1 (a)  •  Q1 (b)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  

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Question 01 (a)
Show that $\displaystyle (3!)^{n}$ divides $\displaystyle (3n)!$ for every positive integer n.

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
Remember that $\displaystyle 3!=6$. An induction argument works or a simple counting argument will suffice.

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Solution 1

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 problem is equivalent to showing that $\displaystyle 6^{n}$ divides $\displaystyle (3n)!$. So we must show that $\displaystyle (3n)!$ contains n copies of both 2 and 3. Now, notice that in $\displaystyle (2n)!<(3n)!$ we have that $\displaystyle (2n)!$ contains 2n consecutive numbers of which n are even. Hence $\displaystyle 2^{n}\mid (2n)!\mid (3n)!$. Now, notice that $\displaystyle (3n)!$ contains 3n consecutive numbers and so n of these must be divisible by 3 (one in every 3 consecutive numbers is divisible by 3. Hence $\displaystyle 3^{n}\mid (3n)!$ and combining these results (since 2 and 3 are coprime) gives us our result.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For a proof by induction, notice that $\displaystyle (3!)^{1}=6\mid 6=(3n)!$. Assume that $\displaystyle 6^{n}=(3!)^{n}\mid (3n)!$. For $\displaystyle n+1$, we have $\displaystyle (3!)^{n+1}=6\cdot 6^{n}$. Now, $\displaystyle 6^{n}\mid (3n)!$ and $\displaystyle (3(n+1))!=(3n+3)!=(3n+3)(3n+2)(3n+1)(3n)!$. Thus, it suffices to show that $\displaystyle 6\mid (3n+3)(3n+2)(3n+1)$. Clearly $\displaystyle 3\mid (3n+3)$ and since this a product of three consecutive numbers, one must be even and so 2 also divides this product. Thus $\displaystyle 6^{n+1}\mid (3n+3)!$ and this completes the proof.

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