Evaluate Fraction: 30^4+324 to 78^4+324

[anemone]

Evaluate, without the aid of a calculator, of the following fraction:

$\dfrac{(30^4+324)(42^4+324)(54^4+324)(66^4+324)(78^4+324)}{(24^4+324)(36^4+324)(48^4+324)(60^4+324)(72^4+324)}$.

I hope you will find this problem interesting, if it's not too difficult or intriguing to solve for.

 

[Saitama]

Evaluate, without the aid of a calculator, of the following fraction:

$\dfrac{(30^4+324)(42^4+324)(54^4+324)(66^4+324)(78^4+324)}{(24^4+324)(36^4+324)(48^4+324)(60^4+324)(72^4+324)}$.

I hope you will find this problem interesting, if it's not too difficult or intriguing to solve for.

Spoiler

From Sophie Germain identity:
$$x^4+4\cdot 3^4=(x^2+2\cdot 3^2-2\cdot x\cdot 3)(x^2+2\cdot 3^2+2\cdot x\cdot 3)=(x(x-6)+18)(x(x+6)+18)$$
In this problem $x=6k$ i.e
$$(x(x-6)+18)(x(x+6)+18)=18^2(2k(k-1)+1)(2k(k+1)+1)$$
For numerator, $k$ has the values $5,7,9,11,13$ and for denominator, $4,6,8,10,12$. Hence, the fraction is:
$$\frac{(2(5)(4)+1)(2(5)(6)+1)(2(7)(6)+1)(2(7)(8)+1)(2(9)(8)+1)(2(9)(10)+1)\cdots (2(13)(14)+1)}{(2(4)(3)+1)(2(4)(5)+1)(2(6)(5)+1)(2(6)(7)+1)(2(8)(7)+1)(2(8)(9)+1)\cdots 2(12)(13)+1)}$$
Most of the terms cancel and we are left with:
$$\frac{2(13)(14)+1}{2(4)(3)+1}=\frac{365}{25}=\boxed{\dfrac{73}{5}}$$

 

[anemone]

Spoiler

From Sophie Germain identity:
$$x^4+4\cdot 3^4=(x^2+2\cdot 3^2-2\cdot x\cdot 3)(x^2+2\cdot 3^2+2\cdot x\cdot 3)=(x(x-6)+18)(x(x+6)+18)$$
In this problem $x=6k$ i.e
$$(x(x-6)+18)(x(x+6)+18)=18^2(2k(k-1)+1)(2k(k+1)+1)$$
For numerator, $k$ has the values $5,7,9,11,13$ and for denominator, $4,6,8,10,12$. Hence, the fraction is:
$$\frac{(2(5)(4)+1)(2(5)(6)+1)(2(7)(6)+1)(2(7)(8)+1)(2(9)(8)+1)(2(9)(10)+1)\cdots (2(13)(14)+1)}{(2(4)(3)+1)(2(4)(5)+1)(2(6)(5)+1)(2(6)(7)+1)(2(8)(7)+1)(2(8)(9)+1)\cdots 2(12)(13)+1)}$$
Most of the terms cancel and we are left with:
$$\frac{2(13)(14)+1}{2(4)(3)+1}=\frac{365}{25}=\boxed{\dfrac{73}{5}}$$

Well done, Pranav! :) See, I mentioned that this problem isn't too difficult and I'm happy that you saw the trick to solve this challenge problem! (Happy)

 

[kaliprasad]

Spoiler

we have
$(x^4 + 18^2) = (x^4 + 2 * 18 x^2+ 18^2) - 36 x^2$
= $(x^2 + 18)^2 - (6x)^2$
=$ (x^2 + 6x + 18)(x^2 - 6x +18)= (x (x+6) + 18)(x(x-6) + 18)$
So
$30^4 + 324 = ( 30 * 36 + 18)(30 * 24 + 18)$
$24^4 + 324 = ( 30 * 24 + 18)( 18 * 24 + 18)$

by expanding numerator and denominator we are left with( as other terms cancel)

value = $\frac{(78 * 84 + 18)}{(24 * 18 + 18)}$
= $\frac{18 (13 * 28 + 1)}{18 * (24 + 1)}$
= $\frac{(13 * 28 + 1)}{ (24 + 1)}$
= $\frac{365}{25}$
= $\frac{73}{5}$

 

[DreamWeaver]

Great thread, and great solutions too! (Bow)

 

[anemone]

Great thread, and great solutions too! (Bow)

Thank you for your kind words, DreamWeaver!:) I really appreciate it!