MHB Solve the Puzzle: 4, 5, 14, 185, .... - Ray Salmon

  • Thread starter Thread starter karush
  • Start date Start date
AI Thread Summary
The discussion revolves around solving the sequence 4, 5, 14, 185, with participants exploring various mathematical approaches. Initial observations include differences between the numbers and their factors, leading to polynomial equations that can generate the sequence. One participant proposes that the next number could be derived from a polynomial function, suggesting that the result can be arbitrary. Another participant identifies a pattern where each number is 11 less than a perfect square, concluding that the next number in the sequence is 34,214. The conversation highlights the complexity and flexibility in generating sequences from limited data.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
ok somebody sent me this on youtube but I don't think is a workable series

Rimbus Jift
@ray salmon Quick IQ test... Solve: 4, 5, 14, 185, ...

so first the difference between the numbers is 1,9,171so far just some ideas
The factors of 171 are 1, 3, 9, 19, 57, 171
$9^0 =1$ and $4+1=5$
$9^1 =9$ and $5+9=14$

$9+5=14$
$9+14=19$
$9\cdot 19=171 $
anyway ?

i plugged into W|F but didn't return a series
 
Last edited:
Mathematics news on Phys.org
karush said:
ok somebody sent me this on youtube but I don't think is a workable series

Rimbus Jift
@ray salmon Quick IQ test... Solve: 4, 5, 14, 185, ...

so first the difference between the numbers is 1,9,171so far just some ideas
The factors of 171 are 1, 3, 9, 19, 57, 171
$9^0 =1$ and $4+1=5$
$9^1 =9$ and $5+9=14$

$9+5=14$
$9+14=19$
$9\cdot 19=171 $
anyway ?

i plugged into W|F but didn't return a series
You should know by now that you can choose any number to be the next one. For example we can use
[math]f(x) = -28 x^4 + \dfrac{917}{3} x^3 - 1130 x^2 + \dfrac{5014}{3} x - 815[/math]

and get f(1) = 4, f(2) = 5, f(3) = 14, f(4) = 185. and the next number in the series will be f(5) = 0.

-Dan
 
ok actually I haven't seen that,
the few series I worked on just plug and played with values of n till you got an eq to generate the series
the imperative "solve" does not insist that it is series generated by an eq but I assume that was the intention
why woufd f(5)=0 or is that just arbitrary

Anyway it does seem slam dung stuff
 
karush said:
ok actually I haven't seen that,
the few series I worked on just plug and played with values of n till you got an eq to generate the series
the imperative "solve" does not insist that it is series generated by an eq but I assume that was the intention
why woufd f(5)=0 or is that just arbitrary

Anyway it does seem slam dung stuff
f(5) is arbitrary. For example:
[math]f(x) = -\dfrac{671}{24} x^4 + \dfrac{1221}{4} x^3 - \dfrac{27085}{24} x^2 + \dfrac{6677}{4} x -814[/math]

gives f(1) = 4, f(2) = 5, f(3) = 14, f(4) = 185, and f(5) = 1.

etc. And you can do other fits aside from polynomials pretty much so long as you have 5 unknowns and the system can be solved. (Polynomials are easy to fit which I why I prefer to use them for demonstrations.)

A problem like this assumes that you can figure out a pattern but unfortunately any more information that you might get could change that answer. So I feel that problems like this are just silly.

-Dan
 
ok I think so too ,...

but curious
what online series calculators are good if you just give a list of 6 numbers which assumes a generator eq
I guess W|F will but haven't tried
 
Someone just posted this on YT again in a random Covid video (9/10/23). I started down the same line as above, noticing the differences of each number in the series is a multiple of 3, but this didn't lead anywhere. I played with the numbers and realized they are each 11 away from a square.

In fact 4²-11 = 5; 5²-11 = 14; 14²-11=185. So the answer is 185²-11 = 34,214.
 
  • Like
Likes Nik_2213 and PeroK
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
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...
Back
Top