Show that $(13)(14)(27)/6$ is an integer without calculating it or factorizing&canceling any of the numbers in the parenthesis Our teacher gave us a very difficult problem which is to me very very unfamiliar 
Show that $(13)(14)(27)/6$ is an integer without calculating it or factorizing&canceling any of the numbers in the parenthesis
I don't know how it can be shown without factorization 
 A: $\begin{eqnarray}{\bf Hint}\quad f(n) &=& (n\!+\!1)\,n\,(2n\!+\!1)\\ &\,=\,& (n\!+\!1)\,n\,(3n\, -\, (n\!-\!1))\\
&=& 6n \dfrac{(n+1)n}2 - 6\dfrac{(n+1)n(n-1)}{3\cdot 2}\\
 &=&  \color{#c00}6\,n{n\!+\!1\choose 2}\, -\, \color{#c00}6\,{n\!+\!1\choose 3}\quad {\bf QED}\end{eqnarray}$ 
Alternatively $\ \color{#0a0}{f(n)-f(n\!-\!1)} = \color{}6\, n^2\,$ so $\,6\mid f(n)\,$ follows by induction, since $\,6\mid f(0) = 0,\,$ and $\,\color{}6\mid \color{blue}{f(n\!-\!1)}\,\Rightarrow\,\color{#c0}6\mid f(n)=\color{#0a0}{f(n)\!-\!f(n\!-\!1)}+\color{blue}{f(n\!-\!1)}, \,$ yields the induction step.
Remark $\ $ Summing $\,{f(n)-f(n\!-\!1)} = 6 n^2$ yields telescopic cancellation giving $\ f(n) = \color{#c00}6 \sum n^2.\,$ This proof by telescopic induction does not require any prior knowledge of the closed-form for the sum (as in the accepted answer). This is a prototypical example of the great power of telescopy in inductive proofs. Many other examples can be found in my posts on telescopy.
A: $$13 \equiv 1 \pmod{6}$$
$$14 \equiv 2 \pmod{6}$$
$$27 \equiv 3 \pmod{6}$$
Now simply apply the fact that $a \equiv b \pmod{n}$ and $x \equiv y \pmod{n} \Longrightarrow$ $ax \equiv by \pmod{n}$:
$$(13)(14)(27) \equiv 1\cdot2\cdot3 \equiv 0 \pmod{6}$$
A: Hint: $$\begin{align*}
\dfrac{(13)(14)(27)}{6}&=\dfrac{\color{blue}{13}(\color{blue}{13}+1)(\color{blue}{13}{\cdot2}+1)}{6} \\ &\color{grey}{\text{and:}}\,\,\,\sum_{i=1}^ni^2=\dfrac{n(n+1)(2n+1)}{6}
\end{align*}$$
