$(k-1)(k)(k+1)$ are three consecutive numbers so they will be divisible by $3! = 6$ AND $(k-1)(k)(k+1)$ is multiplied by 8 so the product will be divisible by $8 \times 6$ which is $48$.

It is proved now that $8(k-1)(k)(k+1)$ and $48k$ are both divisible by 48 and when you add two numbers which are divisible by 48, obviously the sum will also be divisible by 48.

Let us assume $n$ to be an even number $2k$. Now, the equation becomes- $2k(4k^{2}+20)$ and then $8k^{3}+40k$. Taking $8k$ to be common like this- $8k(k^{2} + 5)$, we can split "5" in the equation to form this- $8k(k^{2} + 6 - 1)$. And factorizing $k^{2} + 1$ gives us, $8k((k+1)(k-1) + 6)$. Here we would be using an identity that the product of $n$ consecutive numbers is divisible by $n!$. Thus in this case we have 3 consecutive numbers- $k, k+1, k-1$, which means that this part is divisible by $6$. So, this part- $8 * (k) * (k+1) * (k-1)$ is divisible by $48$ (as we have 8 multiplied to it and 6 divides it). Now the remaining part is- $8k*6$ which in itself is divisible by 48. Thus the sum of 2 numbers divisible by 48 will give a number divisible by 48.

Q.E.D (Hope if you Like it!). And thanks to @MehulArora for his initial solution. Cheers!

Taking $n=2x$ and simplifying a bit gives,$8(x^3+5x)$.Now,we just need to prove that $x^3+5x\equiv 0\pmod{6}$.For doing that we first write $x(x^2+5)\\
=x(x^2+6-1)\\
=x^3+6x-x\\
=x^3-x+6x\\
x(x^2-1)+6x\\
(x-1)(x)(x+1)+6x$.Now,we just need to prove that $(x-1)(x)(x+1)\equiv 0\pmod{6}$ which is obvious.Hence proved.@Vatsalya Tandon

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestSince n is even, Let n=2k for some k.

Then, The Expression becomes,

$2k({(2k)}^{2}+20)$ = $8k({k}^{2}+5)$

Thus, 8 is a factor. Now, we know that one of k or ${k}^{2}+5$ has to be a multiple of 2. Hence We get that 2 is also a factor.

let $n({n}^{2}+20)= {n}^{3}+20n$.

One can see that, ${n}^{3}=n (mod 3)$. So, ${n}^{3}+20n$ = n+20n =21n Which is congruent to 0 (mod 3)

Hence, We find that 3 is also a factor. So, The number is completely divisible by $8*2*3=48$.

Hence, Proved :D

@Vatsalya Tandon .

This was my first Proof. If you spot any mistakes, Kindly inform it to me. I will clean it up.

Log in to reply

Cheers!!! xD

Log in to reply

LOL, Cheers! xD

Log in to reply

Log in to reply

Log in to reply

Let n be 2k :

Then, $2k(4k^{2} +20)$

$8k^{3}+40k$

$8k(k^{2}+5)$

$8k(k^{2}-1+6)$

$8k((k+1)(k-1)+6)$

$8(k-1)(k)(k+1)+8k \times 6$

$8(k-1)(k)(k+1)+48k$

Now, $48k$ is clearly divisible by 48 and,

$(k-1)(k)(k+1)$ are three consecutive numbers so they will be divisible by $3! = 6$ AND $(k-1)(k)(k+1)$ is multiplied by 8 so the product will be divisible by $8 \times 6$ which is $48$.

It is proved now that $8(k-1)(k)(k+1)$ and $48k$ are both divisible by 48 and when you add two numbers which are divisible by 48, obviously the sum will also be divisible by 48.

Log in to reply

Yeah even i did the same way

Log in to reply

Sorry Guys! Updated the Values! Sorry for the Inconvenience!

Log in to reply

Ok I Finally Get This-

Let us assume $n$ to be an even number $2k$. Now, the equation becomes- $2k(4k^{2}+20)$ and then $8k^{3}+40k$. Taking $8k$ to be common like this- $8k(k^{2} + 5)$, we can split "5" in the equation to form this- $8k(k^{2} + 6 - 1)$. And factorizing $k^{2} + 1$ gives us, $8k((k+1)(k-1) + 6)$. Here we would be using an identity that the product of $n$ consecutive numbers is divisible by $n!$. Thus in this case we have 3 consecutive numbers- $k, k+1, k-1$, which means that this part is divisible by $6$. So, this part- $8 * (k) * (k+1) * (k-1)$ is divisible by $48$ (as we have 8 multiplied to it and 6 divides it). Now the remaining part is- $8k*6$ which in itself is divisible by 48. Thus the sum of 2 numbers divisible by 48 will give a number divisible by 48.

Q.E.D (Hope if you Like it!). And thanks to @MehulArora for his initial solution. Cheers!

Log in to reply

My Pleasure @Vatsalya Tandon

Log in to reply

Taking $n=2x$ and simplifying a bit gives,$8(x^3+5x)$.Now,we just need to prove that $x^3+5x\equiv 0\pmod{6}$.For doing that we first write $x(x^2+5)\\ =x(x^2+6-1)\\ =x^3+6x-x\\ =x^3-x+6x\\ x(x^2-1)+6x\\ (x-1)(x)(x+1)+6x$.Now,we just need to prove that $(x-1)(x)(x+1)\equiv 0\pmod{6}$ which is obvious.Hence proved.@Vatsalya Tandon

Log in to reply