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Posts posted by bonanova


I'll give these a shot.
Spoiler x=10a; 2a=8; x=40
 Beggar is a girl.
 50.

Fifth man was driving a car.
He originally outpaced the others until they ran and caught up at their destination.  80 = 60+20.

He took a marble with his closed fist while pointing to the jar for his pick.
He then exposed a black marble in his hand, meaning his pick was a blue marble.  You will hang me.
 Move the cat and come back. Move the mouse and retrieve the cat. Move the dog and come back. Move the cat.
 Suffocation. Two goldfish.
 No smoke from an electric train.
 For sleeping on the job.
 No. Fuel weighing more than the bird had been consumed before it landed on the truck.

Hi Tomson, and welcome to the Den.
SpoilerNot much at all.
An oldie but goodie.

Poe, E. Near a Raven
Midnights so dreary, tired and weary.
Silently pondering volumes extolling all bynow obsolete lore.
During my rather long nap  the weirdest tap!
An ominous vibrating sound disturbing my chamber's antedoor.
'This,' I whispered quietly, 'I ignore.' Mike Keith, 1995

I watch word riddles in awe from the sidelines cuz my brain is wired deductively rather than inductively.
On this one, tho, as a musician of sorts, I have to say, bravo!
Great riddle and great solve! 
OK thx.
SpoilerThen I'm guessing it's a lateral puzzle. Thinking ...

15 hours ago, Wilson said:That's the answer I'm looking for. If you put your two posts together, you've got it.
SpoilerFF?

What does FF? mean?
It clearly means "FF" is the question.

Seriously. What does "FF" mean?
It's a hexadecimal number.
Hex numbers have digits { 0 1 2 3 4 5 6 7 8 9 A B C D E F } with values 015 and
binary representations { 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 }.
That said, FF can be expressed as 1111 1111.

I'm waiting for something useful
OK. Now we have to do some reverse engineering.
Recall that 2 (hex) is 0010 and B (hex) is 1011?

*Tapping fingers*
Almost done.
In binary logic, "not" changes 1 (true) to 0 (false) and vice versa.
And "A or B" is true if either of A or B is true.

Seriously dude. This was a word riddle?
OK I could have posted it in the logic puzzle thread.
But you solved it, didn't you?

If you say so, but I don't see
2B = 0010 1011.
Not 2B = 1101 0100.
(2B) or (not 2B) = 1111 1111 = FF.
That, as we have just shown, is the question.
And Willie S. just might be a writer.
Very nice solve.

What does FF? mean?

4 hours ago, Thalia said:I think the inclusion of the word "who" indicates it applies to the American. Can't be sure without clarification though.
Agree.
If "Heidi" is the intended subject of "found," a 2nd "who" would serve no purpose except to confuse. It would not be used. Instead,
Heidi is { immediately right of (Person A) } and found { more caches than (Person B) }.
But since there are two "who"s, it's proper to bind them to the same (and closest) antecedent:
..... (Person A) { who has flown ... } and { who found more ... }
7) Heidi is immediately right of the person who has flown over from the States, and who found more caches than the UK cacher. 40 Traditional cache where found by a European cacher, their favourite type.
19 hours ago, rocdocmac said:Agree. They could not be Pitches 3 and 5, for example.
I also fixed a typo in my interpretation post.

Clarification:
SpoilerDo you agree the pieces at { e1 c3 g3 e5 } form a square?

SpoilerDoes this have anything to do with Ramsey theory?

My interpretations
Spoiler
Country(Pitch2) = France
Fav(USA) = wherigos.

Pitch(Marie) + 1 = Pitch(Chile)
Fav(Marie) = multi
Finds(Marie) = 2 Finds(Spain)
Fav(Spain) /= letterbox.

Fav(Heidi) = mystery
Finds(Heidi) > Finds(Hugo) > Finds(Frank)

Contry(Bridgette) = Germany
Pitch(Bridgette) = Pitch(Frank)  2
Pitch(Frank) /= 7
Pitch(UK)  Pitch(Bridgette) > 1

 Pitch(Chile)  Pitch(Belgium)  /= 1 (*) EDIT: should read /= 2
Pitch(Chile) > Pitch(Stanley)
Pitch(earth) = 1
(*) Interpret "next door but one" to mean exactly one intervening pitch.
Clue 5 then says this does not apply to Chile and Belgium.

Pitch(Juan) = 3
ManX = { Hugo Juan Stanley }
Finds(Manx) = 16
Pitch(Manx) < Pitch(Frank)

Pitch(Heidi) = Pitch(USA) + 1
Finds(USA) > Finds(UK)
Finds(Fav = traditional) = 40
Country(40 = Finds) = { Belgium France Germany Spain UK }
I believe gives this solution
Spoiler Stanley UK earth 2
 Marie France multi 10
 Juan Chile letterbox 16
 Bridgette Germany traditional 40
 Hugo Spain virtual 5
 Frank USA whereigo 4
 Heidi Belgium mystery 8

Country(Pitch2) = France

SpoilerMy shadow?

Yes, sir, that is exactly the right track.

On 5/27/2018 at 2:38 AM, CSIQ said:... the girl/boy, boy/girl options are, in fact, only one option.
They are equally likely but they are not the same.
But it's worse than that. The OP is deficient, because it does not tell us how we came to know what we know.SpoilerSuppose the person that is telling us about Teanchi and Beanchi has first decided this: If a family has at least one son, I will say the family "has at least one boy." If not, then I will say the family "has at least one girl." But let's say we don't know that is his algorithm for describing families to us. Which is true, because the OP does not disclose it. But what would happen in that case?
We know all twochild families belong, in equal numbers, to these four gender groups: { GG GB BG BB }. If T&B are GB BG or BB, our informant will say T&B have at least one boy. Only in the GG case, where there is not "at least one son," will he say T&B have at least one girl. The probability that the other kid is a girl is then 100%!
Or suppose our informant has first decided this: If the elder of two children is a son, I will say the family "has at least one boy." If not, then I will say the family "has at least one girl." What happens in that case? If T&B are BG or BB, he will say T&B have at least one boy. Otherwise, when T&B are GG or GB, he will say T&B have at least one girl. The probability that the other kid is a girl is then 50%.
Or suppose our informant has decided this: If there is at least one daughter, I will say "the family has at least one girl." Otherwise, I will say "the family has at least one boy." Now, if T&B are GG GB BG he will say T&B have at least one girl, and the probability that the other kid is a girl is 1/3.
Ouch. That is the case that seems most likely, but we can't assume our informant's algorithm with certainty. The OP as stated is deficient.
Instead, let's create a situation where we know how we know what we know, and therefore will let us find the probability that "the other kid is a girl," unambiguously.
SpoilerIf births are genderneutral, we know that families with exactly two children divide equally into four categories for the sex of their older/younger child. Namely: { GG GB BG BB }. The idea now is to eliminate the last category, but do it in a way that does not inject any bias. If we can do that, then we can say GG is one of three equally likely cases { GG GB BG } so the answer is 1/3.
Here's one way.
Pick a number from a hat containing all number from 1100, and say it turns out to be 13. Now assemble thousands and thousands of twochildren families from the general population. This means equal numbers of GG GB BG and BB families. Now, in some random order, we ask them this question: "Is at least one of your children a girl?" and let's say that the 13th family to answer "Yes" is the Smith family.
We can be certain that the Smith family is type GG GB or BG with equal probability.
Doing that, we know the answer is 1/3.

So, while not being a correct solution, this would meet that qualification?
Spoiler{ George Bush NoChild LeftBehind }

Once letters start getting removed it quickly gets easier. At the start it's very hard. How much of a clue are you willing to share? I don't want to disclose too much, but would you be willing to confirm the politician is male and contemporaneous? Initials or country of affiliation might be too revealing.

This sounds a lot, but not exactly, like eliminating variables from sets of equations. Is that the idea?


7 hours ago, Donald Cartmill said:Capt Eds' answer is not quite as straight forward as could be.
Both answers state correctly that (a) winning distances give speed ratios and (b) combined speed ratio gives combined winning distance. What part of that can be more (or less) straightforward?




At the annual Brain Denizen picnic, there were the inevitable games, among them the everpopular threelegged race. Three teams were formed by tying one contestant's right leg to another' s left leg. Fortunately all six contestants made it to the finish line without any broken bones! For purposes of this puzzle we assume all three teams ran the 100meter course at constant speed. Team 2, comprising BMAD and Thalia, were able to beat Team 3, comprising rocdocmac and DejMar by 20 meters, but lost to the winner, Team 1, comprising plasmid and plainglazed, by 20 meters as well. By how many meters did Team 1 beat Team 3?

Spoilerx^{2} = C
x = +/ C^{1/2}
x^{3} = +/ C^{3/2}

On 5/31/2018 at 9:10 AM, DejMar said:Very nice!
There is another solution that ends two moves earlier, leaving pegs only on the edges.
13 BRAIN TEASERS TO MAKE YOU THINK HARD AND TO BOOST YOUR BRAIN POWER
in New Logic/Math Puzzles
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1. Five.
2. If he were the fastest and slowest, there would be 1 runner. Add 49 in front and behind and you have 99 runners.
3. Clues 1 and 2 eliminate 0, and Clue 4 eliminates 8. So Clue 1 establishes 2.
Clue 3 says 2 and 6 are correct but wrongly placed. So we have 62.
Clue 2 says ( 1 or 3 ) is correct ( so it's 612 or 632 ) but wrongly placed, so it's 632.
4. 3 sons, 4 daughters.
5. Ninth.
6. Nine. Or Ten.
7. Bruce and Frank are dead. Jack is alive.
8. 2 men ($6), 5 women ($7.50), 13 children ($6.50)
9. Six. If they were all women.
10. Push the cork into the bottle. Shake the coin out.
11. (T)(e)(n)(H)(o)(r)(s)(e)(s)
12. Ten minutes
13. $20