MHB Need help to reverse engineer a problem...........................

  • Thread starter Thread starter jayearl
  • Start date Start date
  • Tags Tags
    Engineer Reverse
jayearl
Messages
3
Reaction score
0
I hoping that someone may help me find a formula for this problem...

Starting with a specific, large-sum number (ex. 5,408,286,291),

find 4, three or four digit numbers that, when multiplied, will arrive at this large-sum number. It could
be 3 larger numbers or 4 smaller numbers. It doesn't matter really just as long as we arrive at the
final sum.

Since I'm not a math student, I'm hoping to find a formula that can accomplish this. I guess
this is basically doing multiplication backwards. I know it's probably a strange request but I figured
that someone on this forum would know now.

Thank you
Jayearl
 
Mathematics news on Phys.org
Hello, and welcome to MHB, jayearl!

I've moved your thread because this forum is a better fit for the topic of the thread.

Your best bet here is to use a calculator/computer to factor the number (doing so by hand could be very tedious), for example:

W|A - Factor 5,408,286,291

On that page, we find the prime factorization:

$$5,408,286,291=3^2\cdot61\cdot1049\cdot9391$$

So, if we wanted 4 factors, we could use:

$$5,408,286,291=9\cdot61\cdot1049\cdot9391$$

And if we wanted 3 factors:

$$5,408,286,291=549\cdot1049\cdot9391$$
 
Thank you for the assistance.
But I don't believe I explained myself well.
(again, I'm not a math-guy at all so I don't know where to start)

I would like to have 3 or 4 numbers (each with 3 or 4 digits) that, when multiplied, arrive at a specific sum.
ex. 469 X 102 X 309 X 910 = _______number that I first specify______ OR
2005 X 7311 X 3017 = _______number that I first specify_______

Everyday I'll start with a new 9-digit number. Then, I'll need to work backwards? to find the factors that arrive at this new, daily sum.

Is this more clear?
Thank you again for any assitance.
Jayearl
 
Just to clear up some terminology, when you multiply several numbers together you get a product, not a sum. A sum is the result of addition. :D

There is no easy method/formula for factoring large numbers in general...so the best method is to use a calculator/computer to run known efficient algorithms for spitting out the factorization.
 
I believe this is what I need to know how to do:
"so the best method is to use a calculator/computer to run known efficient algorithms for spitting out the factorization"

This is where I don't know what to do...I know even know what an algorithms is .. OR I would be willing to pay someone to figure this out for me.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top