6
$\begingroup$

How can i show that $$6\mid (n^3+11n)$$

My thoughts: I show that $$2\mid (n^3+11n)$$ $$3\mid (n^3+11n)$$

And $$n^3+11n=n\cdot (n^2+11)$$ And if $n=x\cdot 3$ for all $x \in \mathbb{N}$ then: $$3\mid (n^3+11n)$$ And if not:

The cross sum of$$n^2+11$$ is multiple of 3.

Can this be right or is there a simple trick?

$\endgroup$
18
$\begingroup$

$$n^3+11n=\underbrace{(n-1)n(n+1)}_{\text{Product of three consecutive integers}}+12n$$

See The product of n consecutive integers is divisible by n factorial

OR The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)

| cite | improve this answer | |
$\endgroup$
5
$\begingroup$

It is easier to understand the part for factor $3$ to check if $n^3+11n = 0 \pmod 3$.

Then either $n = 0 \pmod 3$ or $n = \pm 1 \pmod 3$.

The $n^2 + 11$ is then $0 \pmod 3$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Nice approach +1 $\endgroup$ – Shailesh Feb 26 '16 at 11:38
3
$\begingroup$

You can also prove this by induction.

$$(n+1)^3+11(n+1)=(n^3+11n) + 3n(n+1)$$

$6$ divides the first part by induction hypothesis. Then $3$ divides $3$ and $2$ divides $n(n+1)$ (either $n$ or $n+1$ is even) so $6$ divides $3n(n+1)$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

$$ n^3+11n = 12 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3} = 6\left(2 \binom{n}{1} + \binom{n}{2} + \binom{n}{3}\right) $$

This is obtained by using repeated differences and Newton's interpolation formula: $$ \begin{array}{llll} 0 & 12 & 30 & 60 & 108 & \\ 12 & 18 & 30 & 48 & \\ 6 & 12 & 18 & \\ 6 & 6 & \\ 0 & \\ \end{array} $$

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ This technique seems to be a @robjohn favorite. $\endgroup$ – lhf Feb 26 '16 at 11:43
1
$\begingroup$

$$(n+1)(n+2)(n+3) = n^3 + 6n^2 + 11n + 6.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ (Essentially the same as lab bhattacharjee's answer.) $\endgroup$ – Unit Feb 26 '16 at 14:04
  • 1
    $\begingroup$ It may be essentially the same, but +1 -- because it comes more naturally looking at the given expression. $\endgroup$ – Shailesh Feb 26 '16 at 15:52
1
$\begingroup$

By brute force:

$$(n^3+11n)\bmod6\equiv(n\bmod6)^3+11(n\bmod6)\mod6$$ and you can proceed by trying the integers $0$ to $5$.

Mentally, $$0,12,30,60,108,180.$$

You can speed-up the computation by dropping any multiple of $6$ you meet and by using $-n$ instead of $+11n$.


Incrementally:

Compute the delta between two consecutive terms

$$(n+1)^3+11(n+1)-n^3-11n=3n(n+1)+12.$$

As one of $n$ and $n+1$ is certainly even, the delta is divisible by $6$.

The property holds for all $n$, as $0^3+11\cdot0$ is divisible by $6$.

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

Here is a proof by induction,

  1. setting $n=1$, one should get $$n^3+11n=1^3+11\cdot 1=12$$ obviously, the above number $12$ is divisible by $6$ hence statement is true for $n=1$.

  2. Assume that the number $n^3+11n$ is divisible by $6$ for $n=k$ then $$k^3+11k=6\lambda$$or $$k^3=6\lambda-11k\tag 1$$

  3. Now, setting $n=k+1$, $$(k+1)^3+11(k+1)$$$$=k^3+3k^2+14k+12$$ $$=6\lambda-11k+3k^2+14k+12$$ $$=6\lambda+3(k^2+k+4)$$ since, $k^2+k+4$ is even for all integer $k$ hence setting $k^2+k+4=2m$, $$=6\lambda+3(2m)$$ $$=6(\lambda+m)$$ since, $(\lambda +m)$ is an integer hence, the above number $6(\lambda+m)$ is divisible by $6$

Hence, $n^3+11n$ is divisible by $6$ for all integers $n\ge 1$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ OP was asking for a simple trick. $\endgroup$ – Shailesh Feb 26 '16 at 11:36
-1
$\begingroup$

using mathematical induction

crucial step is $$p(n)\Rightarrow p(n+1)$$

Let $$p(n)=n^3+11n=6k$$ then $$p(n+1)=(n+1)^3+11(n+1)=n^3+3n^2+3n+1+11n+11=$$ $$=6k+3n^2+3n+12=6k+6(\frac{n^2+n}{2}+2)=6(k+\frac{n(n+1)}{2}+2)$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I ask downvoter what is wrong with my answer! $\endgroup$ – Adi Dani Jul 30 '16 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.