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How to Solve It?

How can we prove that-

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

[Given : n is a positive even number]

Note by Vatsalya Tandon
2 years, 1 month ago

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Since 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. Mehul Arora · 2 years, 1 month ago

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@Mehul Arora Cheers!!! xD Nihar Mahajan · 2 years, 1 month ago

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@Nihar Mahajan LOL, Cheers! xD Mehul Arora · 2 years, 1 month ago

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@Mehul Arora Good! Parth Lohomi · 2 years, 1 month ago

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@Parth Lohomi Thanks! :D Mehul Arora · 2 years, 1 month ago

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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. Ashtag Gaming · 2 years, 1 month ago

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@Ashtag Gaming Yeah even i did the same way Aditya Kumar · 1 year ago

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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! Vatsalya Tandon · 2 years, 1 month ago

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@Vatsalya Tandon My Pleasure @Vatsalya Tandon Mehul Arora · 2 years, 1 month ago

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Sorry Guys! Updated the Values! Sorry for the Inconvenience! Vatsalya Tandon · 2 years, 1 month ago

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Comment deleted Apr 19, 2015

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Comment deleted Apr 18, 2015

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@Mehul Arora 1 isn't even plz see read the question more carefully. Adarsh Kumar · 2 years, 1 month ago

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@Adarsh Kumar Yeah. Sorry :P :P :P xD Mehul Arora · 2 years, 1 month ago

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@Siddhartha Srivastava Delete it @Siddhartha Srivastava :) -_- Mehul Arora · 2 years, 1 month ago

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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 Adarsh Kumar · 2 years, 1 month ago

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