How to Solve It?

How can we prove that-

n×(n2+20) n \times (n^{2} + 20) is divisible by 48.

[Given : n is a positive even number]

Note by Vatsalya Tandon
4 years, 4 months ago

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

Then, The Expression becomes,

2k((2k)2+20)2k({(2k)}^{2}+20) = 8k(k2+5)8k({k}^{2}+5)

Thus, 8 is a factor. Now, we know that one of k or k2+5{k}^{2}+5 has to be a multiple of 2. Hence We get that 2 is also a factor.

let n(n2+20)=n3+20nn({n}^{2}+20)= {n}^{3}+20n.

One can see that, n3=n(mod3){n}^{3}=n (mod 3). So, n3+20n{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 823=488*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.

Mehul Arora - 4 years, 4 months ago

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Cheers!!! xD

Nihar Mahajan - 4 years, 4 months ago

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LOL, Cheers! xD

Mehul Arora - 4 years, 4 months ago

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@Mehul Arora Good!

Parth Lohomi - 4 years, 4 months ago

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@Parth Lohomi Thanks! :D

Mehul Arora - 4 years, 4 months ago

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Let n be 2k :

Then, 2k(4k2+20) 2k(4k^{2} +20)

8k3+40k8k^{3}+40k

8k(k2+5)8k(k^{2}+5)

8k(k21+6)8k(k^{2}-1+6)

8k((k+1)(k1)+6)8k((k+1)(k-1)+6)

8(k1)(k)(k+1)+8k×68(k-1)(k)(k+1)+8k \times 6

8(k1)(k)(k+1)+48k8(k-1)(k)(k+1)+48k

Now, 48k48k is clearly divisible by 48 and,

(k1)(k)(k+1)(k-1)(k)(k+1) are three consecutive numbers so they will be divisible by 3!=63! = 6 AND (k1)(k)(k+1)(k-1)(k)(k+1) is multiplied by 8 so the product will be divisible by 8×6 8 \times 6 which is 4848.

It is proved now that 8(k1)(k)(k+1)8(k-1)(k)(k+1) and 48k48k 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.

Ashtag Gaming - 4 years, 4 months ago

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Yeah even i did the same way

Aditya Kumar - 3 years, 3 months ago

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

Vatsalya Tandon - 4 years, 4 months ago

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Ok I Finally Get This-

Let us assume nn to be an even number 2k2k. Now, the equation becomes- 2k(4k2+20)2k(4k^{2}+20) and then 8k3+40k8k^{3}+40k. Taking 8k8k to be common like this- 8k(k2+5)8k(k^{2} + 5), we can split "5" in the equation to form this- 8k(k2+61)8k(k^{2} + 6 - 1). And factorizing k2+1k^{2} + 1 gives us, 8k((k+1)(k1)+6)8k((k+1)(k-1) + 6). Here we would be using an identity that the product of nn consecutive numbers is divisible by n!n!. Thus in this case we have 3 consecutive numbers- k,k+1,k1k, k+1, k-1, which means that this part is divisible by 66. So, this part- 8(k)(k+1)(k1)8 * (k) * (k+1) * (k-1) is divisible by 4848 (as we have 8 multiplied to it and 6 divides it). Now the remaining part is- 8k68k*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!

Vatsalya Tandon - 4 years, 4 months ago

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My Pleasure @Vatsalya Tandon

Mehul Arora - 4 years, 4 months ago

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Taking n=2xn=2x and simplifying a bit gives,8(x3+5x)8(x^3+5x).Now,we just need to prove that x3+5x0(mod6)x^3+5x\equiv 0\pmod{6}.For doing that we first write x(x2+5)=x(x2+61)=x3+6xx=x3x+6xx(x21)+6x(x1)(x)(x+1)+6xx(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 (x1)(x)(x+1)0(mod6)(x-1)(x)(x+1)\equiv 0\pmod{6} which is obvious.Hence proved.@Vatsalya Tandon

Adarsh Kumar - 4 years, 4 months ago

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