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Posts posted by bonanova


there are exactly 26 sentences
each misses the letter of that letter in the alphabet
I looked at that possibility in my post above.
Unfortunately,
[1] I count 50 sentences, not 26
[2] The third sentence has a "c" in the word "called"  to wit:
Sentence 1:
I've spent the previous hour or so discovering one interesting property concerning linguistics.
[no a b f j k m q v w x z  11 missing letters]  No "A"
Sentence 2:
It's given me much enjoyment while writing the text you're viewing just now.
[no a b c d f g k p q z  10 missing leters]  No "B"
Sentence 3:
Noted as far in the past as 1853, I speak of a unique literary marvel called the lipogram.
[no b h j w x z  6 missing letters]  Alas, there's a "C"
Is that the way you count the sentences?
It looks like you have two sentences as your first sentence.

OK, this might be it.
Circles [aka bags] under the eyes.

What is the puzzle?
I believe the answer is derived by the symbols and their position.
Like this old one: [ignore the dots  they're there just to get proper spacing]
STAND . . . . . . 2 . . TAKING
 teUnd  
. . I . . . . . . THROW . MY
which means, of course,
I understand you intend to overthrow my undertaking

I got here too late to get my answer in, but here's my reasoning without reading what must be the only solution:
[1] There's enough time to get the job done.
[2] If you do two slices in the first two time slots, you lose  cuz you cant do both sides of the third slice at the same time.
[3] So you have to involve the third slice during the 2nd time slot.
[4] Then there's only one possibility for the 3rd time slot, and it's the answer.

Writersblock, I agree with your comments  absence of j's and q's aren't surprising.
I posted basically my dead ends, throwing my ideas into the mix, tho they were wrong.
I suspect that there is a solution regardless of what are probably typos.
The author most likely paid attention to the critical issues of letters
being present or absent in all the right places.
I kind of lit up when I saw no "a" in 1st sentence and no "b" in sentence 2. c'est la vie.
Should be interesting to learn eventually what the answer is.

i hope these couples are on their way to counseling.
damn wheres the trust??
You mean we have to take a counselor across, too?

Looking for missing letters:
as writersblock has noted, no letter is missing from the entire text.
how about letters missing from individual words?
missing "e" in Shakespeare
missing "o" in choose  in penultimate paragraph
missing "f" in twelfth
These spell "foe" ??
oh wait. missing "t" from rewritten. "foet"?
Letters missing from the initial letters of sentences:
b d e j k p q r u v x z  unremarkable.
How about missing numerical digits?
all digits are mentioned except for 6 and 9.
How about assembling the nth words from the nth sentences?
Sentence 11 has only 10 words.
extra [not missing] "n" in Burmann [as noted by unreality]
Burmann is correct spelling; so Burnmann is probably a typo
Taking clues from the text:
Looking for forced or unnatural expâ€‹ressions or phrases that might be inserted as clues
uses of lipograms "in periods gone" [see next spoiler]
create lipogram from the "sentences" on this page
I tried lettercounting by sentence  no pattern.
I noticed "a" missing from 1st sentence and "b" from 2nd, but there is a "c" in the 3rd.
fairy tale inspired me to try something similar  progressive loss of letters.
This seems like a strong clue. I looked for, but didn't find it.
reflect on Mr [sic] Burnmann [sic] note "period gone" from Mr.
giving special significance to Mr. B perhaps?
I'll look more closely at what is said about Burmann next.
"hides between the words"
this seems significant because it doesn't describe what a lipogram is.
So ... what is hiding between the words?
"suddenly dawns" > ?
"smacks you in the face" > ?
Which might lead to:
if "in periods gone" is significant, try looking only at sentences that end in "!", "?" and possibly ":" sentences with "periods gone".
taken together, they are missing "j" "q" and "z".

A friend insisted the answer "had to be" 50%. So I asked him this:
Of all the couples in the world with 2 children, statistically speaking,
what fraction have a 1 boy and 1 girl?
He answered: one half.
and what fraction have 2 boys and 0 girls?
He answered: one fourth.
So, just taking these couples,
what fraction have 2 boys and 0 girls?
He answered: One third; and then: Oh.... OK I get it now.

watery eyes?

Right ... the pentagon is the stumper. The clue, What if there were five points? makes it come to mind, cuz the points of a pentagon create sides and diagonals of only one type, just like as square does. Then you just remove a point to get a 4dot solution. It's seldom you get one solution to be fundamentally different from all the others.

cpotting, amazing story. You might find this amusing: A mathematician, an engineer and an accountant were asked What is 2+2?never underestimate the ability of accountants to ignore facts and plod on anyway ("Numbers be d***ned  the amounts don't matter, just as long as they balance!").Mathematician replied 4
Engineer replied 4.000 plus or minus 0.001
Accountant replied  What to you want it to be?

Can you share your solution?

My answer is correct, because
I was lying.

How about...
"Do you refuse to answer this question"?
If he answers "Yes," he lies.
If he refuses to answer [attempting an affirmative response] he hasn't answered "Yes".
Melchang, are you saying this question can, at some point be answered yes, according to situations and environments?

Answer for a jet:
Yes.
Answer for a propeller plane:
Yes.
Reasoning:
An airplane takes off when its wings generate a lift force greater than its weight. Lift force is determined by air speed. A backwardrunning treadmill will not impede the plane's forward motion, nor limit its air speed, only make its wheels spin twice as fast!
But what if an airplane's engine drove its wheels? Then the answer would be No. It could not achieve forward motion nor generate lift.
That would be an interesting kind of airplane  after it took off it would become a glider!


Very nice!
There are two steps to get to 36pi.
[1] deducing it doesn't depend on R.
[2] computing the R=3 case.
Step [1] can be done logically or mathematically.
Wordblind absolutely wins the prize for the slickest math.

Verifying Martini's result ...
The order on the hill is always [D W J]
The positions, and Jenkins' distance, at 7 critical points are as follows:
0. Start at the bottom [0 0 0 => 0] then ...
2. Jenkins meets Wife [9.6 14.4 14.4 => 17.6]
3. Jenkins reaches top [10.4 13.6 16 => 19.2]
4. Wife meets the Dog [12 12 14.4 => 20.8]
5. Jenkins meets Wife [11.04 13.44 13.44 => 21.76]
6. Jenkins reaches top [9.75 12.16 16 => 24.32]
7. Jenkins reaches bottom [ 0 0 0 => 40.32]

... the man carry one water melon with him, weighing a perfect 80kg, put down the water melon on the other side, than walk back to the other watermelond, weighing 79kg, and then carry that watermelon back to the other side, weighing a perfect 80kg. This answer is plausible, reasonable, and simple, is it not?
Plausible, reasonable and simple. Compelling, even.
Oh wait. The problem states:
"This bridge capacity is up to 80 kgs only, if you exceed this limit the bridge will fall".
"up to 80 kg only" [an open interval] does not include 80. Therefore "a perfect 80kg" "exceed this limit," and so "the bridge will fall."

That part matters most.interprets the word into a concept. 
CorrectAlso, by defining a plane, I assume we are limited to 2 dimensions? 
If two speak the same language, have access to similar dictionaries and understand rules of grammar and syntax, then we say they can communicate in a literate manner.
The meaning of words is a mixture of the "dictionary meaning" and the personal experience of the speaker and listener. The latter component of meaning leads at times to misunderstanding until the context is communicated and understood. Culture and environment can alter generic meanings of words.
How can we prove a word has a meaning? One approach might be to agree on a common dictionary as the authority. A more pragmatic approach might be to say that if two people are comfortable with the notion that they agree when they discuss something, then the words they used to reach agreement themselves have an agreed meaning.
Is this what you had in mind?

Yup.I assume that is the easy one, and you want us to find another.I didn't want to describe the "found" ones yet  it's fun to find them, as well.

p = pi
D = the diameter of the sphere
d = the diameter of the hole
R = the radius of the sphere (.5D)
r = the radius of the hole (.5d)
H = the height of the hole (end to end)
h = the height of the hole (centre to end = .5H)
After drilling, the remaining ring shape is equivalent to a barrel of unknown width (D), 6" high (H) and with the central cylinder portion removed.
The formula for a barrel with sides bent to the arc of a circle
= pH(2DD + dd) / 12
= (1/12)pH(8RR + 4rr)
cpotting,
kudos for finding all these neat [cap and barrel] formulas.
That's does all of the calculus work.
I found it useful to put everything in terms of R and h.
You can do this by noting that RR = rr + hh [Pythagorus]
V[barrel] = ph [8RR + 4rr] /12 = ph [8RR + 4RR  4hh]/6 = 2ph [RR  hh/3]
V[cylinder] = height x area = Hprr = 2ph [RR  hh]
V[barrel]  V[cylinder] = 2ph [RR  hh/3  RR + hh] = 2ph [2hh/3] = 4phhh/3.
Recalling that h=3,
V[barrel]  V[cylinder] = 36p [a constant].
Looking at your derivation, everything is correct.
If you add like terms in your expression for V,
you'll see that all the rr terms add up to zero.
Your derivative expression is correct, also,
except that you should include the constants (1/12)pH8, etc...
[if y(x) = Cxx, then dy/dx = 2Cx, not just 2x]
You'll see again that the rdependent terms add up to zero,
and the derivative is zero.
OK?
Cross out a number  any number  and I'll tell you what it
in New Logic/Math Puzzles
Posted
I hand you a sheet of paper with the following instructions:
[1] Write down a 6digit number. Say you write 352687
[2] Scramble the digits. Say you come up with 762853
[3] Subtract the smaller from the larger. You get: 762853  352687 = 410166
[4] Cross out one of the digits [but not a zero, cuz it's basically not there anyway.] Say you cross out a 6.
[5] Scramble the remaining digits. Say you get: 61401
[6] Tell me the digits. You say 6 1 4 0 1.
What are the odds that I will say, "you crossed out a 6"?