# Finding the value of a fraction

• songoku
In summary, the problem appears to be a degree 4 polynomial which is the product of two quadratics. Dropping the function notation one can write: ##K^4+4\cdot M^4 = \left(\left(K-M \right)^2+M^2\right)\left(\left( K+M \right)^2+M^2\right)##Use this directly in the original expression, recalling that ##324 = 4\cdot 3^4 ####\dfrac{(61^4 + 324)(73^4 + 324)(85^4 + 324)(97^4 + 324)}{(55^4
songoku
Homework Statement
Find:
$\frac{(61^4 + 324)(73^4 + 324)(85^4 + 324)(97^4 + 324)}{(55^4 + 324)(67^4 +324)(79^4 + 324)(91^4 + 324)}$
Relevant Equations
Not sure
I tried to do like this:

$\frac{((76 - 15)^4 + 324)((76 - 3)^4 + 324)((76 + 9)^4 + 324)((76 + 21)^4 + 324)}{((76 - 21)^4 + 324)((76 - 9)^4 + 324)((76 + 3)^4 + 324)((76 + 15)^4 + 324)}$

Then stucked

Thanks

Can you just use a calculator? Or even just multiply it out by hand and then use long division? Why are you factoring the terms with the ()^4 ? Maybe there is more to the problem statement?

songoku and YoungPhysicist
berkeman said:
Can you just use a calculator? Or even just multiply it out by hand and then use long division? Why are you factoring the terms with the ()^4 ? Maybe there is more to the problem statement?
Calculator is not allowed. I can try to do it by brute force, I am asking here whether there are some methods to do it without brute force.

No particular reason why I factor the terms with power of 4, I just think maybe I need to find some ways to simplify them because those power will result in big numbers.

Nothing more to the problem statement, I have posted everything. It is just one line question.

So the only way is brute force?

Thanks

How about if you figure out what is x if x**4 = 324 and see if that helps.

songoku
Is there some context for this problem? What subject was being discussed where this problem appears?

songoku
songoku said:
Calculator is not allowed. I can try to do it by brute force, I am asking here whether there are some methods to do it without brute force.

No particular reason why I factor the terms with power of 4, I just think maybe I need to find some ways to simplify them because those power will result in big numbers.

Nothing more to the problem statement, I have posted everything. It is just one line question.

So the only way is brute force?

Thanks
No. You need not use brute force.

There is quite a bit of cancelling which occurs.

You seem to have noticed the sequence in the terms which are raised to the power 4. You have recognized that ##55 = 76-21\,, ~ 61 = 76-15\,, ~\dots , ~ 97=76+21 ## .
Those could be written using the expression ##76+6(n- \frac 1 2 ) ##. With this, n = −3 gives 55, n = −2 gives 61, ... ,
n = 4 gives 97 .

You may find it helpful to clean that up somewhat. An expression such as ##6n + 1 ## gives 55, 61, ... , 97 when n takes on the values 9, 10, ... , 16 respectively.

See if you can factor the degree 4 polynomial, ##(6n+1)^4+324##, as the product of two quadratics. (It may be helpful to substitute by using ##x## in place of ##6n##.)

songoku
phinds said:
How about if you figure out what is x if x**4 = 324 and see if that helps.
x4 = 22 . 34, so x = ± 3√2

FactChecker said:
Is there some context for this problem? What subject was being discussed where this problem appears?
The teacher said this is question from math competition (do not know what kind of competition) so I don't know which math chapters this question related to

SammyS said:
No. You need not use brute force.

There is quite a bit of cancelling which occurs.

You seem to have noticed the sequence in the terms which are raised to the power 4. You have recognized that ##55 = 76-21\,, ~ 61 = 76-15\,, ~\dots , ~ 97=76+21 ## .
Those could be written using the expression ##76+6(n- \frac 1 2 ) ##. With this, n = −3 gives 55, n = −2 gives 61, ... ,
n = 4 gives 97 .

You may find it helpful to clean that up somewhat. An expression such as ##6n + 1 ## gives 55, 61, ... , 97 when n takes on the values 9, 10, ... , 16 respectively.

See if you can factor the degree 4 polynomial, ##(6n+1)^4+324##, as the product of two quadratics. (It may be helpful to substitute by using ##x## in place of ##6n##.)
I get the hint

Thank you very much for the help berkeman, phinds, factchecker, sammys

@songoku ,

As it turns out, an expression of the form ##\left(u(n)\right)^4+4\cdot M^4## can be written as the following product.

##\left(\left(u(n)-M \right)^2+M^2\right)\left(\left( u(n)+M \right)^2+M^2\right)##

Upon further reflection −

Dropping the function notation one can write:
##K^4+4\cdot M^4 = \left(\left(K-M \right)^2+M^2\right)\left(\left( K+M \right)^2+M^2\right)##

Use this directly in the original expression, recalling that ##324 = 4\cdot 3^4 ##

##\dfrac{(61^4 + 324)(73^4 + 324)(85^4 + 324)(97^4 + 324)}{(55^4 + 324)(67^4 +324)(79^4 + 324)(91^4 + 324)} ##

##= \dfrac{((61-3)^2 + 9)((61+3)^2 + 9)\dots((97-3)^2 + 9)((97+3)^2 + 9)}{((55-3)^2 + 9)((55+3)^2 + 9)\dots((91-3)^2 + 9)((91+3)^2 + 9)} ##

##= \dfrac{((58)^2 + 9)((64)^2 + 9)\dots((94)^2 + 9)((100)^2 + 9)}{((52)^2 + 9)((58)^2 + 9)\dots((88)^2 + 9)((94)^2 + 9)} ##

Noticing the significant amount of "cancellation" which occurs, we are left with a rational expression which is reasonably easy to compute − especially considering the size of the numbers involved in doing a direct computation.

Last edited:
songoku
SammyS said:
@songoku ,

As it turns out, an expression of the form ##\left(u(n)\right)^4+4\cdot M^4## can be written as the following product.

##\left(\left(u(n)-M \right)^2+M^2\right)\left(\left( u(n)+M \right)^2+M^2\right)##

Upon further reflection −

Dropping the function notation one can write:
##K^4+4\cdot M^4 = \left(\left(K-M \right)^2+M^2\right)\left(\left( K+M \right)^2+M^2\right)##

Use this directly in the original expression, recalling that ##324 = 4\cdot 3^4 ##

##\dfrac{(61^4 + 324)(73^4 + 324)(85^4 + 324)(97^4 + 324)}{(55^4 + 324)(67^4 +324)(79^4 + 324)(91^4 + 324)} ##

##= \dfrac{((61-3)^2 + 9)((61+3)^2 + 9)\dots((97-3)^2 + 9)((97+3)^2 + 9)}{((55-3)^2 + 9)((55+3)^2 + 9)\dots((91-3)^2 + 9)((91+3)^2 + 9)} ##

##= \dfrac{((58)^2 + 9)((64)^2 + 9)\dots((94)^2 + 9)((100)^2 + 9)}{((52)^2 + 9)((58)^2 + 9)\dots((88)^2 + 9)((94)^2 + 9)} ##

Noticing the significant amount of "cancellation" which occurs, we are left with a rational expression which is reasonably easy to compute − especially considering the size of the numbers involved in doing a direct computation.

Wow this is just brilliant. What I did was a lot more complex compare to this one. Thank you very much SammyS

## 1. What is a fraction?

A fraction is a number that represents a part of a whole. It is written in the form of one number (numerator) over another (denominator), separated by a line. For example, 3/4 represents 3 parts out of 4 equal parts.

## 2. How do you find the value of a fraction?

To find the value of a fraction, you need to divide the numerator by the denominator. For example, to find the value of 3/4, you would divide 3 by 4, which equals 0.75.

## 3. Can fractions have decimals in them?

Yes, fractions can have decimals in them. These types of fractions are called decimal fractions. For example, 0.5 is a decimal fraction and can also be written as 1/2.

## 4. What is an equivalent fraction?

An equivalent fraction is a fraction that represents the same value as another fraction, even though it may look different. This can be achieved by multiplying or dividing both the numerator and denominator by the same number. For example, 2/4 is equivalent to 1/2 because both fractions represent half of the whole.

## 5. How can we simplify a fraction?

To simplify a fraction, we need to find the greatest common factor (GCF) of the numerator and denominator, and then divide both numbers by the GCF. This will result in a fraction that is in its simplest form. For example, to simplify 4/8, we would divide both numbers by 4 (the GCF), resulting in 1/2.

• Precalculus Mathematics Homework Help
Replies
12
Views
1K
• General Math
Replies
5
Views
1K
• Precalculus Mathematics Homework Help
Replies
1
Views
1K
• Precalculus Mathematics Homework Help
Replies
8
Views
1K
• Precalculus Mathematics Homework Help
Replies
7
Views
2K
• Precalculus Mathematics Homework Help
Replies
7
Views
1K
• Optics
Replies
1
Views
1K
• Precalculus Mathematics Homework Help
Replies
29
Views
2K
• Precalculus Mathematics Homework Help
Replies
14
Views
1K
• Precalculus Mathematics Homework Help
Replies
1
Views
1K