Prove by induction that $$(n+1)(n+2)(n+3)(n+4)(n+5)$$ is divisible by $120$.
I tried to solve it, but could do so only till the inductive step. I assumed: $$p(k)=(k+1)(k+2)(k+3)(k+4)(k+5), and $$ $$p(k+1)=(k+2)(k+3)...(k+6)$$ Then, I distributed it as: $$p(k+1)=(k+2)(k+3)...(k+5)(k+1+5) \\=(k+1)(k+2)...(k+5)+5(k+2)(k+3)...(k+5)\\=p(k)+5(k+2)(k+3)...(k+5).$$ I got stuck over here. Am I right till here? Can someone tell what to do next?
If you insist on doing it by induction, look at $$p(k+1)-p(k)=5(k+2)(k+3)(k+4)(k+5).$$ Among the four consecutive numbers $k+2,k+3,k+4,k+5$, one number is a multiple of four, and another (different) number is a multiple of two. There is also at least one multiple of three. So $5(k+2)(k+3)(k+4)(k+5)$ is divisible by $5\times4\times2\times3=120$, which implies the result.
With very little adjustment, you can rephrase this argument without induction, which is probably the cleaner way to do it.
=== new answer ====
Actually it is easy to prove or more general case of $(n+1)(n+2) ..... (n+k)$ is divisible by $k!$.
And this is a case of proving the induction step by .... induction. A proof within a proof as it were....
Base case: $n+1$ is divisible $1! = 1$. Always.
Induction step: Assume $(n+1)..... (n+(k-1))$ is divisible by $(k-1)!$ for any $n$ then we will prove $(n+1)....(n+k)$ is divisible by $k!$ for any $n$. And we will prove it by induction!
Base case: $1*2*......*k = k!$.
Induction step: Suppose $(n+1) ......(n+k)$ is divisible by $k!$.
then $(n+2)......(n+k)(n+k+1) =$
$(n+2)......(n+k)([n+1]+k)=$
$(n+2)......(n+k)[n+1] + k(n+2)......(n+k)=$
$\color{blue}{(n+1)(n+2)......(n+k)} + k\color{red}{(n+2)......(n+k)}$.
Now we began by assuming $\color{blue}{(n+1)(n+2)......(n+k)}$ is divisible by $k!$.
And $\color{red}{(n+2)......(n+k)}$ is the product of $k-1$ consecutive numbers. And we are doing this entire proof within a prook under the assumption that the product of any $k-1$ consectualve numbers is divisible by $(k-1)!$. So $\color{red}{(n+2)......(n+k)}$ is divisible by $(k-1)!$ and $k\color{red}{(n+2)......(n+k)}$ is divisible by $k*(k-1)! = k!$.
So $\color{blue}{(n+1)(n+2)......(n+k)} + k\color{red}{(n+2)......(n+k)}$ is the sum of two numbers divisible by $k!$ so is itself divisible by $k!$.
That proves the induction step of our proof within a proof, and so proves our proof within a proof: If the product of any $k-1$ consecutive numbers is divisible by $(k-1)!$ then the product of any $k$ consecutive number is divisible by $k!$.
And that proof within a proof was itself the induction step for our claim $(n+1)(n+2).....(n+k)$ will be divisible by $k!$.
....
And so $(n+1)(n+2)(n+3)(n+4)(n+5)$ is divisible by $5! = 120$.
===old answer===
If $(n+1)(n+2)(n+3)(n+4)(n+5) = 120k$ then
$(n+2)(n+3)(n+4)(n+5)(n+6) = [(n+2)(n+3)(n+4)(n+5)]([n+1] + 5)=$
$[(n+1)(n+2)(n+3)(n+4)(n+5)] + 5[(n+2)(n+3)(n+4)(n+5)]=$
$120 k + 5\color{blue}{(n+2)(n+3)(n+4)(n+5)}$
So it remains to show that $(n+2)(n+3)(n+4)(n+5)$ is divisible by $24$.
Claim: $(n+1)(n+2)(n+3)(n+4)$ is divisible by $24$.
Induction: If $(n+1)(n+2)(n+3)(n+4) = 24m$ then
$(n+2)(n+3)(n+4)(n+5) = (n+2)(n+3)(n+4)[(n+1) + 4] = $
$(n+1)(n+2)(n+3)(n+4) + 4(n+2)(n+3)(n+4)=$
$24m + 4\color{orange}{(n+2)(n+3)(n+4)}$
So it remains to show $(n+2)(n+3)(n+4)$ is divisible by $6$.
Claim: $(n+1)(n+2)(n+3)$ is divisble by $6$.
Induction: If $(n+1)(n+2)(n+3) = 6j$ then
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Okay....
Do it by induction
First prove (by induction) that $(n+1)(n+2)$ is always divisible by $2$.
(Easy: Base case: $1*2$ is divisible by $2$. Induction: If $(n+1)(n+2) = 2m$ then $(n+2)(n+3) = (n+2)(n+1) + 2(n+2) = 2m + 2(n+2)$ is divisible by $2$.
Then prove by induction that $(n+1)(n+2)(n+3)$ is always divisible by $6$.
Base case: $1*2*3=6$. Induction; if $(n+1)(n+2)(n+3) = 6k$ then $(n+2)(n+3)(n+4) = (n+1)(n+2)(n+3) + 3(n+2)(n+3) = 6k + 3(n+2)(n+3)$ and $(n+2)(n+3)$ is divisible by $2$ so $3(n+2)(n+3)$ is divisible by $6$.
Then prove by induction that any four consecutive .....
....
Look..... just prove any product of $k$ consecutive numbers are divisible by $k!$.
Base case: Any product of $1$ consecutive number is divisible by $1! = 1$.
Induction. If any the product of any $k$ consecutive integers is divisible by $k!$ then we can prove any product of any $k+1$ consecutive integers is divisble by $(k+1)!$ by induction.
Base case: $1*2*....*(k+1) =(k+1)!$
Induction. If $(n + 1)( n+ 2) ...... (n+k+1)$ is divisible by $(k+1)!$ then
$(n+2)(n+2)(n+3) ...... (n+k+1)(n + k+ 2) = $
$(n+2)(n+2)(n+3) ...... (n+k+1)(n + 1+(k+1)) = $
$(n+1)(n+2)....(n+k)(n+k+1) + (k+1)(n+2)(n+3).....(n+k+1)$.
And $(n+1)(n+2)....(n+k)(n+k+1)$ is divisible by $(k+1)!$ and $(n+2)(n+3).....(n+k+1)$ is the product of $k$ consecutive ingtegers so it's divisible by $k!$. So $(k+1)(n+2)(n+3).....(n+k+1)$ is divisible by $(k+1)k!=(k+1)!$.
Let $D$ the operator on $\mathbb{Z}[x]$ given by $P\to D(P)(x)=P(x+1)$. It is easy to see that ($I$=identity) $D(P)-P=(D-I)(P)$ is of degree strictly less than $P$. Hence if $P$ is of degree $s$, $(D-I)^{s+1}=0$ ; in your case, $P$ is of degree $5$, hence $(D-I)^6(P)=0$, and you can see that $P(n+6)$ is a combination with constant coefficients in $\mathbb{Z}$ of $P(n),...P(n+5)$. Now your induction is easy (if for $m, m+1,m+2,m+3,m+4,m+5$ you have the good property, show it works for $m+1,...,m+6$) (note that you can begin with $-5,-4,-3,-2,-1, 0$)
$120=2\cdot 3\cdot 4\cdot 5$. $\Rightarrow 120|1\cdot 2\cdot 3\cdot 4\cdot 5$ and $120|2\cdot 3\cdot 4\cdot 5\cdot 6$.
Now, if $120|k(k+1)(k+2)(k+3)(k+4)$ and $120|(k+1)(k+2)(k+3)(k+4)(k+5)$, then,
$120|(k+1)(k+2)(k+3)(k+4)\{(k+5)-k\}$
$\Rightarrow 24|(k+1)(k+2)(k+3)(k+4)$
Now, use normal induction and proceed.
