Riddle Help

Airb,

It's got me stuck, too!

Welcome to a2k - stick around, see what's here and maybe some clever person will show us the answer.

KP

Thx for the welcome. It's an interesting site so I've added it to my list of "daily's". I actually did get the solution to the previous riddle...but not to spoil anyone's fun I'll leave it unsolved for now and add another one (no solution on this one yet).

A strong variation:
1,3,11,67, ?

Enter the next number:

-Airb

Well it doesn't look like there are any others who had an answer for the first riddle so I'll post the answer.

multiply each term by (3^n+1)/2.

I must admit I did a poor job of relaying the riddle since the title does give at least one hint "Math, three brains deep".

Any thoughts to the second one? Still looking for a solution to the "A strong variation riddle".

That's one riddle I actually spend over 5 minutes on now. I really can't think of the answer.

hehe, Craven, remember what i said that one time when you discussed sozobe's riddle? Any number will do!

By the way the link to what i talked about is
http://www.able2know.com/forums/viewtopic.php?t=6026
I'll be thinking about a more "reasonable" solution for this one.

TGR,

I can think of many many nubers that I can come up with a connection for but none that I feel are the answer (which I believe is mathematic or related to language).

But yes, I agree, with sequence riddles there's almost always more than one answer. Heck with most riddles there is.

Now my next question is did you have my hamster avatar first? If so I'll change mine.

Well thanks for taking a look guys, I do appreciate it. And TechnoGuy I agree that there are certainly multiple answers to this riddle....but some are more correcter than others

Well here's a solution to the Strong Variant

n^(n-1) + (n-1)= x
n= 1,2,3,4,5,6 ...

1^(1-1) + (1-1)=1

2^(2-1) + (2-1)=3

3^(3-1) + (3-1)=11

4^(4-1) + (4-1)=67

5^(5-1) + (5-1)=629

6^(6-1) + (6-1)= 7781

As TGR pointed out there is certainly more than one formula to produce this series but this one works for me. Thx again.

1, 2, 10, 140, 5740, ?
1 8 130 5600 ? <- It doesnt seem to be an arithmetic sequence

We could try the geometric sequence (a*r^n) with n = 0 or 1 in the first term
1, 2, 10, 140, 5740

a*r^0 = 1
a*r^1 = 2
...
this means a = 1, and r = 2, which makes no sense. Now let's try n = 1

a*r^1 = 1
a*r^2 = 2
...
Here we obtain r = 2, and a = 1/2, which also doesnt work. Since this function rises real fast, it must be of a high degree. Let's try to factor out a number of each number in the sequence. The factors of 5740 are 2,2, 5, 7, 41 (I used my brand new TI-92+ ), the factors of 140 are 2,2,5,7, the factors of 10 are 2, 5, the factors of 2 are 2. See a pattern here? (41 is a prime)

2 = 2 = 2
2,5 = 2 + 5 = 7
2,5,7 = 2 + 5 + 7 = 14
2,5,7,41 = 2 + 5 + 7 + 41 = 55
2,5,7,41,?

Do you see how 2+5+7 = 14? 14 is the inverse with digits of 41? 2+5 = 7, and 7 is already it's own inverse. 2+5+7+41 = 55, and 55 is it's own inverse. I conclude that the next time is a factor of 2,5,7,41,55, (keeping in mind we will need 2 2's), the next number is 2*2*2*5*7*41*55 = 631,400. 1 has no factors of 2, 10 has 1, 140 has 2, 5760 has 2, so i figured the next 3 have 3 factors of 2, the next 4 after that have 4, etc.

Completing this sequence, it looks something like this:
1,2,10,140,5760,631400,6945400. Sorry for the previous dead ends, but it helps me do the problem. (I don't solve it on paper but right on here, if I can).

Hope that this helps, Airb

BTW, Craven you can try to figure out the solution with the arithmetic sequence. Here is what I found so far.

1 = 1
8 = 2*2*2 or 2^3
130 = 2*5*13
5600 = 2*2*2*2*2*5*5*7 or 2^5*5^2*7

Maybe someone can explain, how Airb get the answer for 1st riddle like this:
1,2,10,140,5740, ?
multiply each term by (3^n+1)/2? And why you look at that hint:
"Math, three brains deep"

The answer to the riddle is as follows...

1 | * 2
2 | * 5 (2 + 3)
10 | * 14 (5 + 3*3)
140 | * 41 (14 + 3*3*3)
5740 | * 122 (41 * 3*3*3*3)
= 700280

Thats both of those challenges aree on http://hackquest.com
tut tut
Your supposed to work them out by your self.
Not to spoil the fun or any thing