+0  
 
0
1781
3
avatar

8=56 7=42 6=30 5=20 3=?

Guest Mar 27, 2015

Best Answer 

 #3
avatar
+5

It turns out that the '?' can be whatever we want it to be!

 

Allow me to illustrate. Take for example, f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42, and evaluate f(8), f(7), f(6), f(5), f(3). It turns out that:

 

f(8)=56,
f(7)=42, 
f(6)=30, 
f(5)=20, 
f(3)=9.

 

Wait, what sorcery is this? Turns out that although the popular rule f(x)=x(x-1), which gives f(x)=6, satisfies the known values in the sequence, that f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42 also satisfies them--except with a different value of f(3)!

 

Here's another one that also works but gives f(3)=12:
f(x)=(1/20)x^4-(13/10)x^3+(271/20)x^2-(543/10)x+84

 

And here is one where f(3)=π
f(x)=(1/120)(π-6)x^4-(13/60)(π-6)x^3+(1/120)(251π-1386)x^2+(1/60)(3138-533π)x+14(π-6)

 

Finally, in general if you want the '?'=k, i.e., f(3)=k where k is the value of your choice, then
(1/120)(k-6)x^4-(13/60)(k-6)x^3+(1/120)(251k-1386)x^2+(1/60)(3138-533k)x+14(k-6)

 

More details here:
https://www.scribd.com/doc/260182194/Elementary-Sequences

Guest Jan 5, 2016
 #1
avatar
+5

The answer is 6.

Guest Mar 27, 2015
 #2
avatar+27128 
+5

Possibly as follows:

7x8 = 56

6x7 = 42

5x6 = 30

4x5 = 20

3x4 = 12

.

Alan  Mar 27, 2015
 #3
avatar
+5
Best Answer

It turns out that the '?' can be whatever we want it to be!

 

Allow me to illustrate. Take for example, f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42, and evaluate f(8), f(7), f(6), f(5), f(3). It turns out that:

 

f(8)=56,
f(7)=42, 
f(6)=30, 
f(5)=20, 
f(3)=9.

 

Wait, what sorcery is this? Turns out that although the popular rule f(x)=x(x-1), which gives f(x)=6, satisfies the known values in the sequence, that f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42 also satisfies them--except with a different value of f(3)!

 

Here's another one that also works but gives f(3)=12:
f(x)=(1/20)x^4-(13/10)x^3+(271/20)x^2-(543/10)x+84

 

And here is one where f(3)=π
f(x)=(1/120)(π-6)x^4-(13/60)(π-6)x^3+(1/120)(251π-1386)x^2+(1/60)(3138-533π)x+14(π-6)

 

Finally, in general if you want the '?'=k, i.e., f(3)=k where k is the value of your choice, then
(1/120)(k-6)x^4-(13/60)(k-6)x^3+(1/120)(251k-1386)x^2+(1/60)(3138-533k)x+14(k-6)

 

More details here:
https://www.scribd.com/doc/260182194/Elementary-Sequences

Guest Jan 5, 2016

31 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.