May be a Simple Series Puzzle !

[Commander]

What comes next in the Series :

 

 

There may be many Logical Solutions. But to make it limited

we declare that it is a serial function where

 

f (1) = 0 ; f (2) = 7 ; f (3) = 28 ; f (4) = 69 and ? = f (5)

 

What is the Function and What should be the value of f (5)

 

Any Function which satisfies the first four numbers ie 0 , 7 , 28 , 69

for f (1) , f (2) , f (3) , f (4) will be accepted !

Edited August 29, 2015 by Commander

[John Cuthber]

For n = any integer greater than 73 the answer is n

 

You need to multiply out (x-0) (x-7) (x-28)( x-73) (x-n) to give a 5th order polynomial

The function is then the roots of that polynomial in ascending order.

[Commander]

For n = any integer greater than 73 the answer is n

 

You need to multiply out (x-0) (x-7) (x-28)( x-73) (x-n) to give a 5th order polynomial

The function is then the roots of that polynomial in ascending order.

 

Sorry John there was an error in my Puzzle which I have corrected.

 

Please reconcile !

[John Cuthber]

It doesn't alter my point.

If you have a finite sequence of i(monotonically increasing) integers then they are always the roots of a polynomial in order of increasing size.

And I can put any number I like at the end and choose an equation so that I can justify that choice of number.

 

There are an infinite number of solutions to the problem you have put forward.

[Commander]

It doesn't alter my point.

If you have a finite sequence of i(monotonically increasing) integers then they are always the roots of a polynomial in order of increasing size.

And I can put any number I like at the end and choose an equation so that I can justify that choice of number.

 

There are an infinite number of solutions to the problem you have put forward.

 

OK, then do give a number for the fifth slot and the Polynomial Equation which holds.

 

Then if needed , I can give what number I get on that fifth spot and you can give a Polynomial for it,

 

BTW I think you are on the RIGHT TRACK !

It doesn't alter my point.

If you have a finite sequence of i(monotonically increasing) integers then they are always the roots of a polynomial in order of increasing size.

And I can put any number I like at the end and choose an equation so that I can justify that choice of number.

 

There are an infinite number of solutions to the problem you have put forward.

 

We are not talking about listing the roots of the Polynomial Equations in order which ends the series.

 

This series will go on for ever though it suffices for you to find the fifth element and when you give the Polynomial it will prove that you can derive every further member of the series.

 

That means A B C D E F etc which stand for one positive integer each like 0 7 28 69 ? ? etc

and A = f (x) with x = 1 and f (x) being the Polynomial.

 

Similarly f(x) takes values of B C D E etc where f(2) = B ; f(3) = C ; f(4) = D ; f(5) = E etc ....

[John Cuthber]

You have missed my point.

I can pick any number bigger than 69 and write out the polynomial trivially.

If you add a dozen more numbers I can still write another expression for the "next " number to be anything I like as long as it's bigger than all the ones you have put forward.

I suspect the number you expect is about 130 (not to give the game away) but I think you should check your arithmetic again.

[Commander]

You have missed my point.

I can pick any number bigger than 69 and write out the polynomial trivially.

If you add a dozen more numbers I can still write another expression for the "next " number to be anything I like as long as it's bigger than all the ones you have put forward.

I suspect the number you expect is about 130 (not to give the game away) but I think you should check your arithmetic again.

 

I am not missing any point.

 

You might be right and have solved the puzzle perhaps.

 

Just tell what is the number and don't worry about giving clues to others. You be the first to solve it.

 

And state what is the Polynomial function f(x) which satisfies the puzzle.

 

I think you can do it soon !

[John Cuthber]

Rather than risk spoiling it I will pretend that I think the answer is 500

OK

So the polynomial is

(x-0) (x-7) (x-28) (x-69) (x-500)

And that polynomial obviously is equal to zero when x takes the values 0,7,28,69 and 500.

For example, at x =7 the second bracketed expression (x-0) is equal to zero- so the overall calculation is equal to zero because it doesn't matter what you multiply it by- the outcome will still be zero.

 

But I could have copied the numbers from the puzzle into the brackets whatever numbers they were.

And I could have put any number I wanted into the last bracket.

 

So, there are always an infinite number of solutions to this sort of puzzle.

And I still wonder if you need to check the arithmetic

1³ = 1 = 0 + 1

2³ = 8 = 7 + 1

3³ = 27 = 28 -1

4³ = 64 = 69 - 5

but I might have got completely the wrong idea.

[Commander]

Rather than risk spoiling it I will pretend that I think the answer is 500

OK

So the polynomial is

(x-0) (x-7) (x-28) (x-69) (x-500)

And that polynomial obviously is equal to zero when x takes the values 0,7,28,69 and 500.

For example, at x =7 the second bracketed expression (x-0) is equal to zero- so the overall calculation is equal to zero because it doesn't matter what you multiply it by- the outcome will still be zero.

 

But I could have copied the numbers from the puzzle into the brackets whatever numbers they were.

And I could have put any number I wanted into the last bracket.

 

So, there are always an infinite number of solutions to this sort of puzzle.

And I still wonder if you need to check the arithmetic

1³ = 1 = 0 + 1

2³ = 8 = 7 + 1

3³ = 27 = 28 -1

4³ = 64 = 69 - 5

but I might have got completely the wrong idea.

 

Well if the Polynomial is :

 

f(x) = (x-0) (x-7) (x-28) (x-69) (x-500)

 

then f(1) = (1-0) (1-7) (1-28) (1-69) (1-500)

= (1) (-6) (-27) (-68) (-499)

which is obviously not = 0

 

Therefore f(1) is not = 0 and the puzzle is not solved by this Polynomial !

 

May be with this new input you can design the Polynomial to solve it.

 

You can do it !

Edited August 30, 2015 by Commander

[John Cuthber]

The numbers are the solutions to the polynomial, not th e results from calculating it.

[Commander]

The numbers are the solutions to the polynomial, not th e results from calculating it.

 

If I had asked what comes after 0 . 4 , 18 , 48 , ?

 

and specify that f(1) = 0 ; f(2) = 4 ; f(3) = 18 ; f(4) = 48

 

Then the answer can be 100

 

and the Polynomial f(x) can be f(x) = x³ - x²

 

Similar Polynomial will be the solution to the Puzzle in the context !

 

It is not very complicated !

[fiveworlds]

0,7,28,69,70 = x^5 - 174x^4 + 9891x^3 - 196294x^2 + 946680x

 

This is also achievable by

 

f(x)=(x+x+x)(x-1)+(10)(x-2)(x-1) - (4.5)(x-3)(x-2)(x-1) - (38/24) (x-4)(x-3)(x-2)(x-1)

 

There is no correct answers to most iq tests if you look at my function it is basically a series of if statements which only execute when the number is above the specified value here (x-3)(x-2)(x-1) it is 3. So really I could just use if statements it would be faster.

Edited September 9, 2015 by fiveworlds

[Commander]

0,7,28,69,70 = x^5 - 174x^4 + 9891x^3 - 196294x^2 + 946680x

 

This is also achievable by

 

f(x)=(x+x+x)(x-1)+(10)(x-2)(x-1) - (4.5)(x-3)(x-2)(x-1) - (38/24) (x-4)(x-3)(x-2)(x-1)

 

There is no correct answers to most iq tests if you look at my function it is basically a series of if statements which only execute when the number is above the specified value here (x-3)(x-2)(x-1) it is 3. So really I could just use if statements it would be faster.

 

Hi,

 

The first function you have given is :

 

f(x) = x^5 - 174x^4 + 9891x^3 - 196294x^2 + 946680x

= x⁵ - 174 x⁴ + 9891 x³ - 196294 x² + 946680 x

 

I think the result for f(1) is

 

x^5 - 174 x^4 + 9891 x^3 - 196294 x^2 + 946680 x where x = 1

= 760104 whereas it should be 0

 

Similarly f(2) becomes

x^5 - 174 x^4 + 9891 x^3 - 196294 x^2 + 946680 x where x = 2

= 1184560 and not 7 which is expected

 

Please reconcile and work out the function : Good try !

[Commander]

Hi all,

 

No more attempts at resolving this ?

[Lord Antares]

/cut

 

Hello John. Can you explain this for someone who is illiterate in mathematics?

Because this seems like exactly the sort of thing that comes up in an IQ test. If there is a learnable method to solve the puzzle, that would invalidate the point of this type of question in an IQ test, right?

 

I cannot really follow the logic of your mathematics here.

[MentalGrind]

0, 7, 28, 69, 136, 235

 

Simple enough, took me just 15 mins....

 

f(n) = (x-1)(x²+2x-1)

 

f(1) = (1-1)(1+2-1) = 0

f(2) = (2-1)(4+4-1) = 7

f(3) = (3-1)(9+6-1) = 28

f(4) = (4-1)(16+8-1) = 69

f(5) = (5-1)(25+10-1) = 136

f(6) = (6-1)(36+12-1) = 235

.

.

.

best,

Anita

Edited January 28, 2017 by MentalGrind

[Commander]

0, 7, 28, 69, 136

 

Simple enough, took me just 15 mins....

 

f(n) = (x-1)(x²+2x-1)

 

f(1) = (1-1)(1+2-1) = 0

f(2) = (2-1)(4+4-1) = 7

f(3) = (3-1)(9+6-1) = 28

f(4) = (4-1)(16+8-1) = 69

f(5) = (5-1)(25+10-1) = 136

 

best,

Anita

 

Hi Anita,

 

Well done

 

I'll post my notes here soon

 

Edited January 28, 2017 by Commander