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

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The discussion focuses on evaluating the fraction involving powers of integers and a constant, specifically $\dfrac{(30^4+324)(42^4+324)(54^4+324)(66^4+324)(78^4+324)}{(24^4+324)(36^4+324)(48^4+324)(60^4+324)(72^4+324)}$. Participants express enthusiasm about the problem's complexity and the cleverness required to solve it. Positive feedback is shared among users regarding the solutions provided. The problem is deemed interesting and manageable, encouraging further engagement. Overall, the thread highlights a collaborative effort to tackle a mathematical challenge.
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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.:o
 
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anemone said:
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.:o

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:
$$\frac{2(13)(14)+1}{2(4)(3)+1}=\frac{365}{25}=\boxed{\dfrac{73}{5}}$$
 
Pranav said:
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)
 
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 = $\frac{(78 * 84 + 18)}{(24 * 18 + 18)}$
= $\frac{18 (13 * 28 + 1)}{18 * (24 + 1)}$
= $\frac{(13 * 28 + 1)}{ (24 + 1)}$
= $\frac{365}{25}$
= $\frac{73}{5}$
 
Great thread, and great solutions too! (Bow)
 
DreamWeaver said:
Great thread, and great solutions too! (Bow)

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