• Support PF! Buy your school textbooks, materials and every day products Here!

Finding the value of a fraction

  • Thread starter songoku
  • Start date
  • #1
1,272
29

Homework Statement:

Find:
[itex]
\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)}
[/itex]

Relevant Equations:

Not sure
I tried to do like this:

[itex]
\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)}
[/itex]

Then stucked

Thanks
 

Answers and Replies

  • #2
berkeman
Mentor
56,671
6,566
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?
 
  • #3
1,272
29
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
 
  • #4
phinds
Science Advisor
Insights Author
Gold Member
2019 Award
15,837
5,481
How about if you figure out what is x if x**4 = 324 and see if that helps.
 
  • #5
FactChecker
Science Advisor
Gold Member
5,323
1,906
Is there some context for this problem? What subject was being discussed where this problem appears?
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,238
964
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##.)
 
  • #7
1,272
29
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

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

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
 
  • #8
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,238
964
@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)##

Added in Edit:
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:
  • #9
1,272
29
@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)##

Added in Edit:
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
 

Related Threads for: Finding the value of a fraction

  • Last Post
Replies
2
Views
1K
Replies
3
Views
10K
  • Last Post
Replies
2
Views
2K
Replies
8
Views
6K
Replies
6
Views
4K
Replies
3
Views
724
  • Last Post
Replies
4
Views
2K
Top