#229 - {Official} Star Trek Trivia

[master_q 0]

{Official} Star Trek Trivia #229

Score One For the Team!

 

 

Oh, NO! A math question!

 

12*; 36; 108; 324; . . . . . .

 

What is the next #?

[i want to know what the next five #s are. I should have a total of 5 replies giving a total of 5 answers.]

 

BONUS POINTS: If someone can give me a standard equation for this kind of sequence, then the team will get an additional 3 points!

 

{EDIT}

*LOL. I think I need to take a typing lesson. The numbers listed now should be correct

 

Master Q

StarTrek_Master_Q@yahoo.com

Edited September 28, 2003 by master_q

[Theunicornhunter 2]

12*; 36; 108; 324; . . . . . .

 

What is the next #?

 

972

[Indy 8]

Next number - 2,916

[removed 0]

7th Sequence = 8 748

 

This equation works, but you have to start the variable x at 0 for it to work: (click on spoiler)

 

Click for Spoiler:

equation: n = 12(3^x)

 

ex1.

n = 12(3^0)

n = 12

ex2.

n = 12(3^1)

n = 36

ex3.

n = 12(3^2)

n = 108

ex4.

n = 12(3^3)

n = 324

 

 

Edited September 28, 2003 by bysty

[master_q 0]

7th Sequence = 8 748

 

This equation works, but you have to start the variable x at 0 for it to work: (click on spoiler)

 

Click for Spoiler:

equation: n = 12(3^x)

 

ex1.

n = 12(3^0)

n = 12

ex2.

n = 12(3^1)

n = 36

ex3.

n = 12(3^2)

n = 108

ex4.

n = 12(3^3)

n = 324

 

 

 

equation: n = 12(3^x)

I don't like that equation.

 

I want the first term to be 1 and not 0.

(x should = 1 and give us an answer of 12. When I plug x=2 I should get 36 and so on)

 

So when n = 12 (our first term) I want to be able to plug 1 into the equation to get 12

 

You are close. Just think about it.

 

 

Master Q

StarTrek_Master_Q@yahoo.com

[Captain Bolivar 0]

The next number is 26,244

[Captain Bolivar 0]

Here is a general equation that gives 12 for an x of 1.

n = 12[3^x - {(2)(3^x-1)}]

 

note:

(3^x-1) is NOT (3^x)-1

(3^x-1) IS 3 to the exponent (x-1)

 

SUB IN 1 FOR x

12 = 12[3^1-{(2)(3^1-1)}]

12 = 12[3 - {(2)(3^0)}]

12 = 12[3 - {(2)(1)}]

12 = 12[3 - 2]

12 = 12[1]

12 = 12

 

SUB IN 3 FOR x

108 = 12[3^3 - {(2)(3^3-1)}]

108 = 12[27 - {(2)(3^2)}]

108 = 12[27 - {(2)(9)}]

108 = 12[27 - 18]

108 = 12[9]

108 = 108

[ensign_beedrill 0]

Fifth number - 78,732

[master_q 0]

Here is a general equation that gives 12 for an x of 1.

n = 12[3^x - {(2)(3^x-1)}]

 

note:

(3^x-1) is NOT (3^x)-1

(3^x-1) IS 3 to the exponent (x-1)

 

SUB IN 1 FOR x

12 = 12[3^1-{(2)(3^1-1)}]

  12 = 12[3 - {(2)(3^0)}]

  12 = 12[3 - {(2)(1)}]

  12 = 12[3 - 2]

  12 = 12[1]

  12 = 12

 

SUB IN 3 FOR x

108 = 12[3^3 - {(2)(3^3-1)}]

  108 = 12[27 - {(2)(3^2)}]

  108 = 12[27 - {(2)(9)}]

  108 = 12[27 - 18]

  108 = 12[9]

  108 = 108

The equation seems to work, but I don't like it being that long. In certain cases you can't avoid a long equation, but if you can then try to make it as small as possible. It works and I guess I'll give credit, but I can think of a much smaller equation that would do the same job.

 

Look Bysty's equation. It is not right, but it is close.

n = 12(3^x)

 

n = 12(3^1) = 36

n = 12(3^2) = 108

. . .

 

Bysty said;

This equation works, but you have to start the variable x at 0 for it to work

 

Now think about that! If you have to "start the variable x at 0," then how do you think you can bump that so you can start with 1?

 

 

Master Q

StarTrek_Master_Q@yahoo.com

Edited September 29, 2003 by master_q

[removed 0]

I think I have it.

 

equation: n = 12(3^(x-1))

 

ex1.

n = 12(3^(1-1))

n = 12

ex2.

n = 12(3^(2-1))

n = 36

ex3.

n = 12(3^(3-1)

n = 108

ex4.

n = 12(3^(4-1)

n = 324

 

 

Edited September 29, 2003 by bysty

[Captain Bolivar 0]

That is much nicer! It's sweet and to the point. Well, it's been a few years since I've taken math; i'm not accustomed to thinking in mathematical terms. I'm actually impressed with myself for creating my huge and almost completely redundant equation. Anywho, Well done Bysty!

[removed 0]

Thanks Commander Bolivar. I used your equation to help me out. I hadn't even thought of the whole subtracting an exponent thing. I was thinking a bit too complex. Lol.

[master_q 0]

Good Job Everyone!

+17 Points

 

 

Master Q

StarTrek_Master_Q@yahoo.com

Edited October 2, 2003 by master_q