bonanova

Better descriptions for two of the numbers: from the crowd --> nobody in particular renounce anger --> remove condemnation that should correct one of them. Then, remember the clue, and ...

No it doesn't. All they need to do is burn completely in the specified time.

Yes. The words either begin or end with a number sound.

When Davey and the boys burst into Morty's last night, they found Alex over in the corner, scribbling on a sheet of paper. Wonder what he's up to tonight, said Jaimie. I'm almost finished, came the reply, pour me an O'Doule's and I'll be right over. Now, I know this infinity stuff scares the b'jesus outta you boys, so I promise to stay away from it, at least for tonight. It's a simple puzzle, really. Ya know how some words start or end with the sound of a number? Well here's 16 of them. For example the word GREAT might be represented as [G][R][8]. You get the idea. So there's sixteen numbers here, but when you've found them, that's only part of it. The real puzzle is what to do with them afterward. Ya know how I don't like to give all that many clues, but I will say this: 16 = 4 + 4 + 4 + 4. When you've figured that out, you'll have a single number, and as promised, it won't be infinity. I'd stand on my head before going there again. At least for tonight. Tell me the single number, and you win. Edited to clarify the "number sounds" either begin or end the word.

Hmmm. Well, here's a start

Well it's finally Friday, and no one has won the game, Alex confessed, and it's no wonder. Writersblock even gave ya a reason why ... it just can't be done. And here's another way to show it's impossible. If ya color one third of the holes blue and one third green and one third red this way... you'll see the first move jumps a green peg over a blue peg into the center red hole. Or, blue jumps green into red. Either way, starting with 12 of each color, ya now have 11 blue, 11 green and 13 red. All odd numbers. And any move after that makes an even number of each color. The colors stay all odd or all even until there are no moves left. But ya see, the winning layout has to have only one peg, say it's red. That means there's an odd number [1] of red and an even number [0] of blue and green. And that's just impossible. I suppose it would have been sporting of me to give ya the colored version of the board to start with, but ... heh heh ... i was running out of drink money and ... well, you know...

The OP gives the range as all numbers. As the number of digits in the number increases, the likelihood that it contains a three increases. In fact the fraction of N-digit numbers that contain a 3 is 1 - [.9]**N. Note that [.9]**N becomes 0 at infinite N.

The holes on the board can be colored red, green and blue in such a way that each move involves a hole of each color. Example: blue peg jumps a green peg and lands in a red hole. Continue making all possible moves, and every hole will eventually be colored. In terms of these colors, think about the starting position and a winning end position. Can a series of moves go from the starting position to a winning end position? In terms of colors, what happens on each move?

Gezundheit! I used a bigger board with pennies ... There is a proof [that it can or can't be done] and it's cute.

Since we've nearly completed the task, might be interesting to compare with what others came up with. Picture:[url:1c45f]http://photos1.blogger.com/blogger/4320/817/1600/75%20bands%20answers.0.jpgList:1: Led Zeppelin 2: Gun & Roses 3: B 52s 4: Black Flag 5: White Zombie 6: Smashing Pumpkins 7: Rolling Stones 8: White Snake 9: The Eagles 10: Blind Mellon 11: Sex Pistols 12: Queen 13: Garbage 14: Alice in Chains 15: Matchbox 20 16: The Eels 17: Dinosaur Jr. 18: 50 cent 19: Beach Boys 20: 9 Inch Nails 21: Seal 22: The Dead Kennedy's 23: Madonna 24: The Monkees 25: Ratt 26: Great White 27: The Police 28: The Oranges 29: Radiohead 30: Korn 31: Lemon Heads 32: The Blues Travelers 33: Iron Maidon 34: The Pixies 35: Prince 36: Phish 37: Red Hot Chili Peppers 38: Blur 39: Twisted Sister 40: Cowboy Junkies 41: Gorillaz 42: U2 43: Crowded House 44: Black Crows 45: Cars 46: BeeGees 47: White Stripes 48: Talking Heads 49: Yellowman 50: Kiss 51: Eminem 52: Deep Purple 53: Pink 54: Hole 55: Jewel 56: Green Day 57: Blondie 58: Pavement 59: Spoon 60: Bush 61: Pet Shop Boys 62: Cake 63: Go-Gos 64: Manic Street Performers 65: The Postal Service 66: Cornershop 67: Scissor Sisters 68: The Vines 69: The Doors 70: Men at Work 71: The Cranberries 72: The Band 73: Cypress Hill 74: Skinny Puppy 75: Was

Alex hadn't been seen for a week. And things just weren't the same at Morty's. Jamie said it was cuz he'd lost his last three bets to the geniuses over at Brainden. But Davey opined Alex was not a sore loser. He'll be back, he said. And it was no sooner said than Alex appeared at the door. All smiles. And with a strange looking board under his arm. Take a look, mates, he said, and laid it on the table: What you see here is Alex's new version of the old solitaire game that you play with pegs. There's one empty hole right there in the center, and 36 other holes each with a peg. You jump pegs over neighbors into empty holes, straight ahead, along the lines ya see there, and remove the pegs that are jumped. When ya can't make any more jumps, the game is over. And if there's only one peg left, you've won. Now who'll take my bet that none of ya is smart enough to win? The last peg doesn't have to be in the center hole, mind ya, but there can be only one peg standing at the end. Go ahead. Try a few games to get the idea. Then think it over. If ya take the bet, you've got until Friday to write down a winning game. I'd advise against it, he winked. Cuz if you try and lose, then I get free drinks till Christmas. Ya might be better off just passing it on to the Brainden crowd. Any takers here? Number the holes like this. --------1 2 3 4 -------5 6 7 8 9 ---10 11 12 13 14 15 16 17 18 [19] 20 21 22 ---23 24 25 26 27 28 -----29 30 31 32 33 ------34 35 36 37 and write moves like this: [1] - 6 [12] 19 - [that means: move #1 - peg 6 jumps and removes peg 12 and lands in hole 19] [2] - ....... etc. 35 total jumps and you win. Sign up now if ya want to try, then post yer solution by Friday. Edit: Take a hint from Alex's wink: Prove that the game can be won. Or not.

Or a tree or a building. If trees and buildings could talk...

Which reminds me of the old quote: All of you who believe in telekenesis please raise my hand.

Not bad ...

Sorry, I misread it, both times. Yes that works. Since the OP focuses on using the box that you have, this certainly qualifies as thinking outside the box. Kudos to both you and WB!

Bartender, pour the man a cold lager. Wait. Pour him 31 lagers.

[2], actually - so long as it has four letters. But [1] is intriguing, also. How about four letters becoming soul fretter / tours refelt / ferret louts or fur or let set since there are only four vowels, it's tough to make six words ...

Ah, but the OP specified that ... Note that you cannot send a key in an unlocked box, since it might be copied.

There is a one to one correspondence between the terms in the sequence and the positive integers. The numbers given correspond to the numbers 1-12. No. But this may be helpful: terms 49 and 51 are respectively 58 and 59 terms 62 and 64 are respectively 70 and 73 terms 99 and 101 are respectively 109 and 114

Take four letters and arrange them into six differently spelled words. No proper names, abbreviations, etc ... you know the drill. Just ordinary words you'd find in any dictionary. Use each letter exactly once in each word. Enjoy!

Alex thought hard before going to Morty's last night after losing bets two nights in a row. But go he did, and with an extra swagger, because he had come up with a challenge that he felt sure no one could meet. You know those number series, like 1, 4, 9, 16, 25 ... and the like? he asked, talking to no one in particular. Well all the ones I've seen are like child's play. Last night I come up with some numbers that none of ya here can figure out - not in a month of Sundays. Then grinning he added, But if anyone should be clever enough, I'll buy him drinks for a month. Davey appeared interested and sauntered over. Alex took out a crumpled sheet of paper and handed it to him. On it were scrawled, in Alex's dirty red ink, these numbers: 4, 5, 8, 8, 9, 9, 12, 13, 13, 13, 17, 18, ... Ya see them numbers, do ya? Well, they just go on forever, they do. And if ya figure out what they are, you'll be able to tell me the 50th, 63rd and 100th terms. And that's what it'll take to win. With that, he sauntered over to shoot darts with Jamie - but not before hollering back, Oh, and tell writersblock he's welcome to give it a try, too.

Alex slides another cold one down the bar to WB.

You want to boil a perfect 3-minute egg. For a timer, you have only two 4-minute fuses, that do not burn at an even rate. Can you have breakfast without a runny [or overly hard] egg?

After being done out of a pint of O'Doule's by writersblock last night at Morty's, Alex conjured up a question calculated to get him even. After WB had downed his cool one, Alex proposed a double or nothing puzzle. To the nearest percentage point, he asked, what percentage of all numbers contain at least one 3? For example, 13, 31, 33 and 103 all contain the digit 3 at least once. But 1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, ... well you get the idea ... don't contain the digit 3 even once. Now I know there's a lot of numbers to check, so I'll make it easy for ya, Alex said. I'll give you five multiple choices. That gives you a 20% chance even if you guess, and much better odds if you yer' the genius ya' make yerself out to be. The percent of all numbers containing at least a single "3" is: [1] 10% [2] 13% [3] 33% [4] 67% [5] 100% What was writersblock's choice, and did he win another pint?

Sliding a frosty cold one down the bar to writersblock!