bonanova
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Posts posted by bonanova
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hazel [i would know]
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Precisely in accord with the OP.Encountering intolerant person doesn't mean you are encountering intolerance. I would take intolerance as an implicit quality of a person, not "the situation" created by some circumstances. If you face (i would say feel) intolerance you are not tolerant.A tolerant person encounters a person who possesses the implicit quality of intolerance.
A person who already possesses a quality would hardly be said to encounter it [within himself].
You may say "feel", but the OP said "encounter".
...mumbles to himself: One thing I can't tolerate is people who can't tolerate my perfect English usage ....
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Good one ...
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Apologies for the friggin bad clues.
Here it is.
Sum of 8 2 8 8 => 26
Sum of 2 1 1 1 => 5
Sum of 4 2 8 4 => 18
Sum of 1 4 2 8 => 15
26.5.18.15 => answer
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By encountering people.
you are contradicting yourself in the question. How can you encounter intolerance if you are tolerant at the first place?I am a tolerant person.
What should be my reaction when I encounter intolerance?
Not all people in the world are tolerant people.
A tolerant person might meet a person that is not tolerant.
And it may be impossible to simply walk away.
For example
[a] business colleague
family member
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How would he face intolerance?This might be better set up as a tolerant zealot. A zealot would have a very hard time not advocating for their point of veiw. -
a tick or a thistle.
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I am a tolerant person.
What should be my reaction when I encounter intolerance?
[1] Tolerate it, since I am a tolerant person; thus condoning something I do not believe in.
[2] Not tolerate it, since it goes against my beliefs; thus ceasing to be a tolerant person.
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Clue summary
The title - return from infinity
Alex's promise to stay away from infinity
Alex standing on his head before going to infinity tonight
improved descriptions for 2 of the 16 numbers
grid with the correct 16 numbers
16 = 4 + 4 + 4 + 4 [make four groups of four numbers]
answer is a single-digit number
do something mathematically simple to the numbers in each group [use only one arithmetic function]
then one final step.
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Not bad.((8 / 2) /
* 8 = 42 + 1 + 1 * 1 = 4
4 * 2 - 8 + 4 = 4
8 / ((1 * 4) / 2) = 4
Right track, but doesn't answer to all the clues, including now this one.
Use only one arithmetic function - [the simplest one]
Then there's one final step.
This is fun ...
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And the answer is ...
... He speaks the truth on Tuesdays.
The assertions of Day 1 and Day 3 can't both be true.
Else there would be two truth days: Day 1 and Day 3.
So he cannot lie on all four days mentioned: M T W and F.
The truth day is M T W or F.
The assertions of Day 1 and Day 3 also can't both be false.
Else the truth day would be M or T and also would be W or F, making two truth days.
The truth day is Day 1 or Day 3.
The assertion of Day 2 can't be true.
Else there would be two truth days: Day 2 and Day 1 or Day 3.
Negating Day 2's statement, we see that
Day 2 is M, T, W or F.
Day 2 can't be T or W.
Else Day 1 would be M or T, on which days he claims to lie. A paradox.
Day 2 can't be F.
Else Day 1 would be Th on which he claims to lie M T, and Day 3 would be Sat on which he claims to lie W F.
This makes two truth days: Day 1 or Day 3 [Thu or Sat] and M, T, W or F.
Day 2 is M.
The truth day is Sun or T.
The truth day can't be Sun.
Else, on Day 3 he lies about lying on both W and F, creating two truth days: Sun and W or F.
The truth day is T.
Check:
Day 1 [sun] I lie on M T. [a lie - he speaks truth on T] - OK
Day 2 [Mon] It's Th, Sat or Sun. [a lie - it's M] - OK
Day 3 [Tue] I lie on W F. [the truth - he does lie on those days] - OK
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Maybe ...
Time warp?
Parallel universe?
Sixth Dimension?
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OP meant to imply the answer was a single-digit number.
Here are the 16 numbers you need to find the answer.
Remember to think in groups of four.
8 2 8 8 => ?
2 1 1 1 => ?
4 2 8 4 => ?
1 4 2 8 => ?
Don't place too much emphasis on the fact they're all powers of 2.
It just turns out that the sounds of these numbers commonly occur in words.
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Wow.
Looks interesting.
Now if I just had another life to spend on this.
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A spoiler is buried in the OP.
It may identify the number, but to win you have to find it from the grid results.
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.
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hints
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Guys guys, I think this math stuff is going crazy. You guys are saying it is 100% to contain a number at least one digit 3.To the nearest percentage, 100% of numbers contain at least one digit 3.Well, lets look it this way. What about a number to contain 3 OR 4. According to the argument it is going to be 200%.
No, because a number can contain a 3 AND 4.
That is very funny ... because there are infinite numbers that contains neither 3 nor 4. (Hope no one will ask for an example).
An example is easy, mentioned in a previous post.
The sequence of numbers 1, 11, 111, 1111, 11111, 111111, 1111111, ...
is infinite and contains no 3's or 4's.
Still, the fraction of all numbers that contain a 3, or that contain a 4,
or that contain any specified digit is 1 - [.9]**N where N is the length
of the number. As N -> infinity, [.9]**N -> 0 and the fraction -> 1.
By the same reasoning, the fraction of all numbers that contain all the digits, 0-9, is also 1.
Mind bending, isn't it?
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I know what No Limit and Pot Limit mean, what does
High Stakes mean? i.e., how much of their stake is
a player allowed to bet on each hand?
Are we told each player's starting stake? If it's $1000,
say, then all players save one could be wiped out on
the first deal.
If the stakes are unlimited, so that no one is ever wiped
out, then the single player among the ten with the most
chips is the winner and all others, still having chips, only
fewer, are the losers.
Is that the question? Which has the most chips after 1000
hands? Nobody ever goes All In?
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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 ...
... make four groups of 4 number each.
Do something mathematically simple to the numbers in each group.
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No it doesn't.I think you light one fuse (Fuse A) on both ends and Fuse B on one end simultaneously. Once Fuse A has completely burned to the middle, 2 minutes have passed. Fuse B should have also burned for 2 minutes, so it will have 2 minutes left. Then light the unburned end of Fuse B while it's still burning in the middle and it should burn out in 1 minute, totalling 3 minutes in all. The only question I have is if this solution requires 2 identical fuses of even burn time, or if it will work as stated in the problem, 2 different fuses with uneven burn rates. I'll try to find a mathematical solution and edit laterAll they need to do is burn completely in the specified time.
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Yes.do all of the numbers fit in the start and end cos i have some that are in the middle.so far i have
1derland
2tor
4ever
2sday
Perchlor8
1derous
every1
then some that seem ok but wont fit or dont have numbers at the start/end
pa10t
her0ism
equ8
The words either begin or end with a number sound.
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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.
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Hmmm.
Well, here's a start
Ra [Re?] had 4 children [brood]:
[1] Nut (sky)
[2] Shu
[3] Tefnut (together the Air)
[4] Geb (Earth)
and Thoth, God of the Moon, Magic and Writing [and a handsome devil to boot]
was the god who overcame the curse of Re, allowing Nut to give birth to her five children, with his skill at games.
Working on the rest of it, instead of sleeping tonight.
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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...
Cut Alex's chain .. [or maybe pull it!]
in New Logic/Math Puzzles
Posted
Alex hadn't been to Morty's for a few days, because, it was later
learned, he was home trying to get brainden to come up on his
aging computer. When Jamie called to see if he was OK, he told
Alex it was down for maintenance, and the boys missed him. And
so, even tho a little red-faced [
], Alex trudged in last night.
I've been thinking, he said, about using this old watch chain in my
pocket for this Friday night's poker game. And he held up a tiny
gold chain with 21 links in it.
OOOOOOOOOOOOOOOOOOOOO
The problem is that I want to be able to make smaller groups of
links. Look here, I'll show you what I mean. Suppose I were to
cut the 8th link. And he sketched this picture:
OOOOOOO C OOOOOOOOOOOOO
[making chains of length 7 and 13 links, and a single, cut link]
Then I could make bets of 1, 7, 8, 13, 14, 20 or 21 links.
But I want to be able to bet any number of links, from 1 to 21.
Ever since last Thursday I've been thinking about how to do that
cutting as few links as possible. Now, I'll buy a pint for the first
one of ya that can match my best idea.
Tell me how many links to cut, and which ones.