Riddle Time

TheGreyRonin · Post by **TheGreyRonin** » Thu Dec 14, 2006 1:56 pm

There's a riddle that was posted recently on a website. I haven't gotten the answer myself, so I thought I'd share it with you folks. "If A is two, and B is 3, and F is 8, how high must you count to get to heaven?" Any takers?

Guest · Post by **Guest** » Thu Dec 14, 2006 9:13 pm

That can't possible make since.. you sure thats right? F = 8? Thats.. just not feasible sencable.. arg!! brain fry

breezeetew · Post by **breezeetew** » Fri Dec 15, 2006 1:38 am

Pharfly1 · Post by **Pharfly1** » Fri Dec 15, 2006 3:08 am

I got 64? Maybe my maths off... Ok, my brian has offically been fried... f=8... is there a new letter in the alphabet??? abcde*fg???

KittyCaller · Post by **KittyCaller** » Fri Dec 15, 2006 4:23 am

Hmmm......why do I get the feeling that numbers aren't going to be the answer? Having said that, I've still not got a clue of the answer.

newhermiemami · Post by **newhermiemami** » Fri Dec 15, 2006 5:26 am

I'm with KC...but having said that I have no idea either. lol.

TheGreyRonin · Post by **TheGreyRonin** » Fri Dec 15, 2006 7:53 am

No clue myself, actually. That's all the hint I was provided with.

Guest · Post by **Guest** » Fri Dec 15, 2006 8:01 am

You would have to count forever?

breezeetew · Post by **breezeetew** » Fri Dec 15, 2006 9:39 am

There are 2 ways to ask, and 2 answers...I have heard it for Heaven's Gate, and that is a totally different answer... but I stand by 70 and will explain later.

TheGreyRonin · Post by **TheGreyRonin** » Fri Dec 15, 2006 2:59 pm

Okay. I checked in with all the answers, did the math myself, and... Breezee got it right! Congratulations!

KittyCaller · Post by **KittyCaller** » Fri Dec 15, 2006 3:13 pm

Can you explain?

TheGreyRonin · Post by **TheGreyRonin** » Fri Dec 15, 2006 3:19 pm

Not easily. I'm waiting to see Breezee's version to post the easiest solution. I'm betting hers is easier; mine took a bit.

TheGreyRonin · Post by **TheGreyRonin** » Fri Dec 15, 2006 4:08 pm

Mine involved at least two equations that were redundant. By permission, here's Breezee's solution:If I add one to the count at every vowel this gives a=2 b=3 c=4 d=5 e=7 f=8 g=9 h=10 i=12 j=13, k=14 l=15 m=16 n=17 o=19 p=20 q=21 r=22 s=23 t=24, u=26 v=27 w=28 x=29 y=30 z=31H=10,E=7,A=2,V=27,E=7,N=17. =70 Much simpler and elegant than how I worked it out. I was overthinking it.