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(#5325526) Re: What's next?
Posted by bestgremlin on 1 Jan 2022 at 8:21AM
Still open for someone to solve this:

5 of M9: C B F B ______?


Clue: sports poetry
(#5325498) Re: New Year's Puzzle
Posted by Dragon Jr on 1 Jan 2022 at 6:58AM
Yes, that is correct. I found a different solution.
(2 + 0!) / .2 - 2 = 13.
(#5325491) Re: New Year's Puzzle
Posted by FragileKitty on 1 Jan 2022 at 6:25AM
2 + 0! + (2 / .2)
(#5325447) New Year's Puzzle
Posted by Dragon Jr on 1 Jan 2022 at 3:46AM
Using the digits 2022 in that order, as well as math symbols, form an expression equal to 13. Valid math operations and symbols are: addition, subtraction, multiplication, division, decimal point, powers, roots, factorials (but not double factorial) and parentheses.
(#5307932) Re: What's next?
Posted by MM David littlefair XI on 20 Nov 2021 at 5:38AM
Junction 5
(#5307794) Re: What's next?
Posted by Jools on 19 Nov 2021 at 9:33PM
I don't know much about sports, especially American ones
(#5307786) Re: What's next?
Posted by bestgremlin on 19 Nov 2021 at 8:31PM
It has something to do with sports...
(#5299747) Re: What's next?
Posted by bestgremlin on 29 Oct 2021 at 9:25AM
Perhaps it will help if I rephrase it as:

5 of the M9
(#5299746) Re: What's next?
Posted by bestgremlin on 29 Oct 2021 at 9:24AM
But where does the "5 of M9" come in, if it were just by numbers?
(#5299745) Re: What's next?
Posted by bestgremlin on 29 Oct 2021 at 9:22AM
Yes, "5 of M9" means something.
That is the key to figuring it out.
(#5295051) Re: What's next?
Posted by Jools on 17 Oct 2021 at 10:23AM
Does "5 of M9" mean something?
(#5294862) Re: What's next?
Posted by MM David littlefair XI on 16 Oct 2021 at 11:51PM
IB if you going like the numbers.
A,B C B D,E F B G H I B
    
3,6,9
32,62,92
(#5294777) What's next?
Posted by bestgremlin on 16 Oct 2021 at 4:23PM

5 of M9: C B F B ______?

-- bestgremlin
(#5289852) Music Puzzle
Posted by amthevessel on 5 Oct 2021 at 8:19AM
The idea of what? came to whom?, while he was still composing what else?.
(#5289030) Re: Nine-Pack
Posted by Dragon Jr on 3 Oct 2021 at 2:13PM
You have correctly solved the puzzle.
(#5289028) Re: Nine-Pack
Posted by Jools on 3 Oct 2021 at 2:05PM
1&2: Decorated, as a cake = Ice-d
2&3: Bell sound = d-ing
4&5&6: Orange of the forest = Berg-am-ot
7&8: Son of Seth and Azura = Enos
8&9: Brand of blender = Os-ter
1&4: Variety of lettuce = Ice-berg
2&5: Structure made by rodents = D-am
3&6: Gold bar = Ing-ot
4&7: Ventriloquist Edgar = Bergen
5&8: Minor Hebrew prophet = Am-os
6&9: Playful aquatic animal = Ot-ter

1=ICE
2=D
3=ING
4=BERG
6=A
7=MOT
8=OS
9=TER

BTW, never heard of Oster
(#5289013) Nine-Pack
Posted by Dragon Jr on 3 Oct 2021 at 1:22PM
1&2: Decorated, as a cake
2&3: Bell sound
4&5&6: Orange of the forest
7&8: Son of Seth and Azura
8&9: Brand of blender
1&4: Variety of lettuce
2&5: Structure made by rodents
3&6: Gold bar
4&7: Ventriloquist Edgar
5&8: Minor Hebrew prophet
6&9: Playful aquatic animal
(#5287907) Re: 25 horses
Posted by FragileKitty on 1 Oct 2021 at 2:13PM
Yes, as I explained, having their names (a number) also be their speed (higher is faster) was just for illustrative purposes.

"...and now we only need to find the two fastest horses remaining" in my specific example was 23 and 24.
(#5287906) Re: 25 horses
Posted by Jools on 1 Oct 2021 at 2:05PM
I assumed FragileKitty listed the horses simply in the (reverse) order they won.
I would have listed them by race & position i.e. A1, A2, C3
(#5287904) Re: 25 horses
Posted by Dragon Jr on 1 Oct 2021 at 2:02PM
Actually, the number of each horse is not necessarily its speed. The second and third fastest horses are the two fastest of the 7th race.
As the five horses in that race are the ones numbered 15, 19, 20, 23 and 24, there are four possible choices for second and third fastest.
These are: 15&20, 19&20, 23&24, 20&24.
(#5287903) Re: 25 horses
Posted by Jools on 1 Oct 2021 at 2:01PM
correct
(#5287892) Re: 25 horses
Posted by FragileKitty on 1 Oct 2021 at 1:24PM
It takes 7 races to determine the fastest 3 horses.

Given 25 horses, name them 1 through 25. To make it easier for illustrative purposes, let each horse's name be its "speed" (but in general we don't know this yet).

Hold 5 races with different horses in each race.

Race 1: 1,2,3,4,5
Race 2: 6,7,8,9,10
Race 3: 11,12,13,14,15
Race 4: 16,17,18,19,20
Race 5: 21,22,23,24,25

The top 3 of each race remain in the running (15 total), so we've eliminated 10 horses.

Eliminated: 1,2,6,7,11,12,16,17,21,22

Remaining:

3,4,5
8,9,10
13,14,15
18,19,20
23,24,25

Next we'll race just the 5 winners:

Race 6: 5, 10, 15, 20, 25

This eliminates 5 and 10 (and the horses they beat).

We know now that 25 is fastest, so we can remove it from subsequent races, and now we only need to find the two fastest horses remaining.

This leaves:

14, 15
19, 20
23, 24

We can remove 14, because 20 beat 15, which beat 14. This leaves us with:

15,
19, 20
23, 24

Race 7: 15, 19, 20, 23, 24

23 and 24 win, so they join 25 in the top three.
(#5287715) 25 horses
Posted by Jools on 1 Oct 2021 at 5:57AM
There are 25 horses and you need to identify the fastest 3. You can race up to 5 horses at a time, but you cannot time them.

What is the minimum number of races needed so you can identify the fastest 3 horses and how?
(#5281331) What's next?
Posted by bestgremlin on 13 Sep 2021 at 11:09AM

5 of M9: C B F B ______?


-- bestgremlin
(#5279280) Re: Four Coins: Another puzzle of 12 coins
Posted by Nigel on 8 Sep 2021 at 12:23PM

Let me clarify. There is a solution, whichever way the scales go.
(#5279131) Re: Music Challenge of the Century
Posted by bestgremlin on 8 Sep 2021 at 4:26AM

We have a WINNER! Congrats macBruck

"Got a feeling twenty-one is gonna be a good year..."
(#5278862) Re: Music Challenge of the Century
Posted by Silkwood on 7 Sep 2021 at 2:06PM
Old Time Rock and Roll

Bob Seger
(#5278586) Re: Music Challenge of the Century
Posted by macBruck on 6 Sep 2021 at 8:18PM
1921...The Who
(#5278161) Re: Music Challenge of the Century
Posted by FragileKitty on 6 Sep 2021 at 5:30AM
I was trying to make a U2 song work, but the dates didn't seem to line up.
(#5278151) Re: Music Challenge of the Century
Posted by Sir Face of the GoldTable on 6 Sep 2021 at 4:54AM
The Who maybe - one of the songs from Tommy i think
(#5278132) Re: Music Challenge of the Century
Posted by bestgremlin on 6 Sep 2021 at 3:22AM
What does Johnny B Goode have to do with a century ago?
Please try again.
(#5277588) Re: Music Challenge of the Century
Posted by macBruck on 4 Sep 2021 at 4:39PM
Chuck Berry...Johnny B goode
(#5277519) Music Challenge of the Century
Posted by bestgremlin on 4 Sep 2021 at 12:46PM
A famous rock and roll song's setting was exactly one hundred years ago.

Can you identify the song and who recorded it?
Singing right

-- bestgremlin
(#5273929) Re: Nine-Pack
Posted by Dragon Jr on 27 Aug 2021 at 6:42AM
Yes, that is correct.
(#5273928) Re: Nine-Pack
Posted by Jools on 27 Aug 2021 at 6:35AM
1&2: ___ lantern (Hallowe'en sight): 2 wds. Jack O
2&3: Without obstacles or barriers. o-pen
4&5&6: ___ Falls (Bullwinkle Moose's home). Frost bi-te
7&8: Paradise garden. Ed-en
8&9: Terminus. En-d
1&4: Mythical winter weather guy: 2 wds. Jack Frost
2&5: Sapporo sash. O-bi
3&6: 5-in-a-row board game. Pen-te
4&7: ___ Flakes (cereal). Frost-ed
5&8: Well: French. Bi-en
6&9: Actor Danson. Te-d

1 JACK
2 O
3 PEN
4 FROST
5 BI
6 TE
7 ED
8 EN
9 D
(#5273912) Nine-Pack
Posted by Dragon Jr on 27 Aug 2021 at 5:58AM
1&2: ___ lantern (Hallowe'en sight): 2 wds.
2&3: Without obstacles or barriers
4&5&6: ___ Falls (Bullwinkle Moose's home)
7&8: Paradise garden
8&9: Terminus
1&4: Mythical winter weather guy: 2 wds.
2&5: Sapporo sash
3&6: 5-in-a-row board game
4&7: ___ Flakes (cereal)
5&8: Well: French
6&9: Actor Danson
(#5273645) Re: Four Coins: Another puzzle of 12 coins
Posted by Jools on 26 Aug 2021 at 10:45AM
yes I had thought of 3x 4 coins, but I thought it would take 2x tries to find the group with the fake coin.

But had a rethink. I can do it if they initially balance, but if they don't it takes an extra try

Still figuring out that bit
(#5273571) Re: Four Coins: Another puzzle of 12 coins
Posted by Sir Face of the GoldTable on 26 Aug 2021 at 5:58AM
Three weighings do suffice here.

as I remember the answer to this I'm deliberately holding back, other than to say a solution is definitely possible... and to give you a hint for thoughts:

symmetry...consider splitting the 12 into 3 groups of 4 and start your comparison from there..
(#5273550) Re: Four Coins: Another puzzle of 12 coins
Posted by macBruck on 26 Aug 2021 at 3:37AM
I reckon it can be done, but still working on it!
(#5272368) Re: Four Coins: Another puzzle of 12 coins
Posted by Jools on 23 Aug 2021 at 12:04PM
I can't see it is consistently possible in 3 attempts unless you already know if the defective one is heavier or lighter. You may manage it is 3 attempts if the first attempt balances, but not if it doesn't.

Unless I'm not thinking clearly enough Puzzled
(#5272339) Re: Four Coins: Another puzzle of 12 coins
Posted by Nigel on 23 Aug 2021 at 10:20AM

Well, any attempts, friends?
(#5262652) Re: Four Coins: Another puzzle of 12 coins
Posted by Nigel on 27 Jul 2021 at 3:15AM

Well, I have a similar one, with 12 coins. You have maximum three chances at weighing coins against each other, and then find the defective one, and also say whether it is heavier or lighter.

(The original puzzle, which was put to me more than 40 years ago, was about billiard balls instead of coins).

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