How do I show via direct proof that $k(k+1)(k+2)$ is divisible by $6$. I showed it was divisible by $2$ because at least one of the multiples is even but could not figure out how to show it is divisible by $3$. I tried making $k$ even or odd and substituting $2q$ or $2q+1$ but have not made much progress. Does anyone have any tips as to what direction I should take? Thanks!
Show via direct proof that $k(k+1)(k+2)$ is divisible by $6$.
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elementary-number-theory
discrete-mathematics
divisibility
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0Hint: exactly one of $n$ consecutive numbers is divisible by $n$. – 2017-02-10
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1look at $k+2 \choose k-1$ (if it counts as a direct proof) – 2017-02-10
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1How about three consecutive numbers have the form $3k, 3k+1, 3k+2$? Or $3k-1, 3k, 3k+1$, or finally, $3k-2, 3k-1, 3k$... – 2017-02-10
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0Just relate it to $\Bbb Z_3=\{[0],[1],[2]\}$ – 2017-02-10
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0$k(k+1)(k+2)=6\binom{k+2}{3}$ :P – 2017-02-10
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1Why is at least one of the *factors* even? Why can't they all be odd? Can you modify that to argue that one of the factors must be divisible by 3? Is i possible that $k$ is not divisible by 3? If so then can $k +1$ not be divisible by 3. If so would it then be possible for $k +2 to be divisible by 3? Hint what is the remainder of k/3? What is the remainder of (k+1)/2? What is the remainder of (k+2)/3? – 2017-02-10
2 Answers
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By the division algorithm, $k$ divided by $3$ yields a remainder of $0$, $1$, or $2$. In other words, there are some integers $q,r$ such that $k=3q+r$ where $r=0,1,$ or $2$.
If $r=0$, then $k=3q$ is divisible by $3$. If $r=1$, then $k+2=(3q+1)+2=3(q+1)$ is divisible by $e$. If $r=2$, then $k+1=(3q+2)+1=3(q+1)$ is divisible by $3$. Therefore, in all cases, at least one of $k$, $k+1$, and $k+2$ is divisible by $3$.
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0Thank you so much! Could not ask for a better explanation! – 2017-02-10
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0Glad to help. When you need to prove a "low level" fact about the integers (like at least one of three consecutive integers is divisible by 3), the division algorithm is probably your best tool. :) – 2017-02-10
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0@user6797501 if you like an answer, upvote it. accept the best one. this is how you say thank you. – 2017-02-10
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If $k = 6a+b$, where $0 \le b \le 5$, then
$\begin{array}\\ k(k+1)(k+2) &= k(k^2+3k+2)\\ &= k^3+3k^2+2k\\ &= (6a+b)^3+3(6a+b)^2+2(6a+b)\\ &= (6a)^3+3((6a)^2b+6ab^2)+b^3+3(36a^2+12ab+b^2)+12a+2b\\ &= 6(6^2a^3+18a^2b+ab^218a^2+6ab+2a)+b^3+3b^2+2b\\ &= 6(6^2a^3+18a^2b+ab^218a^2+6ab+2a)+b(b+1)(b+2)\\ \end{array} $
By directly computing $b(b+1)(b+2)$ for $0 \le b \le 5$, the values are all divisible by $6$ (they are $0, 6, 24, 60, 120, 210$).
Therefore, the product is always divisible by $6$, being the sum of two terms each of which is divisible by $6$.