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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

This was my first Proof. If you spot any mistakes, Kindly inform it to me. I will clean it up. · 2 years, 1 month ago

Cheers!!! xD · 2 years, 1 month ago

LOL, Cheers! xD · 2 years, 1 month ago

Good! · 2 years, 1 month ago

Thanks! :D · 2 years, 1 month ago

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. · 2 years, 1 month ago

Yeah even i did the same way · 1 year ago

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! · 2 years, 1 month ago

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

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

Comment deleted Apr 19, 2015

Comment deleted Apr 18, 2015

1 isn't even plz see read the question more carefully. · 2 years, 1 month ago

Yeah. Sorry :P :P :P xD · 2 years, 1 month ago

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

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 · 2 years, 1 month ago