\((k-1)(k)(k+1)\) 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

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## Comments

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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.

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Cheers!!! xD

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LOL, Cheers! xD

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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.

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Yeah even i did the same way

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

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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!

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My Pleasure @Vatsalya Tandon

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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

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