VSP Potpourri

[CrystyB]

YAY! i get to start a topic here! Finally! [j/k]

1. I might start working on it... tomorrow!
5. I can make it in 4... Damn! [edit3]GOT IT! Qf5 ; if g6 then N:h6 with f8Q next; else Q:h7 I GOT A CHESS PUZZLE !!! no i didn't! [/edit3]
7. i get 1/4. I can explain that. Do i need to?
10. Seems impossible to me. Wait! "they both know that THE number is so-and-so"? WHICH number? What do they know about the other one?
12. workING on it. [edit]GOT IT! 168 289.[/edit]

[edit2] Wait! I always wanted the topic to have in its first post [it can't in its title...] a link to the puzzle.[/edit2]


[edit4]8 i already have 2+ answers and there ARE more[edit6]there are exactly 8, i finished the exhaustive search[/edit6]. BTW what house name begins with d anyway? [/edit4]

PS I lost "edit5" by escaping a submit. I can't even remember what was it about ...



[This message has been edited by CrystyB (edited 09-23-2001).]

[Ghost Post]

2. word play i think the answer to it is:
that, that is, is. is that it? it is! not that. that is not, is not. is that it? it is!

[CrystyB]

i actually thought it can be done with only commas (and the only dot at the end)

[planet_buzz]

Ok, I took a stab at movie trivia:
(Key: Kathy Bates KB, Michael Beihn MB, James Caan JC, Sean Connery SC, Cary Elwes CE, Morgan Freeman MF, Ed Harris EH, Winona Ryder WR, Arnold Schwartzenegger Arnold, Alicia Silverstone AS, Christian Slater CS, Uma Thurman UT)

WR --> Heathers --> CS --> Hard Rain --> MF --> Kiss the Girls --> CE --> The Crush --> AS --> Batman --> UT --> The Avengers --> SC --> The Rock --> MB --> Terminator --> Arnold --> Eraser --> JC --> Misery --> KB

(Read as: "WR was the hero in Heathers to CS who was the villian, CS was the hero in Hard Rain to MF who was the villian, etc.")

...but Ed Harris doesn't fit. He can fit as a villian in The Rock, but then MB doesn't have a villian category, and that doesn't go very far. Also, Arnold can be a villian in Batman and Robin, but that doesn't go far either. Anyone want to finish this off for me?

[Memnoch]

3. Cryptics

tapestries
means?

[Jen Aside]

"that that is, is. is that it?" "it is not." "that that is not, is not. is that it?" "it is."

[Jen Aside]

I'm also inclined to think that the answer to the safe's combination (11) is "just about anything." The question is, "What combination do you try?" rather than "What is the correct combination?"

Incidentally, I'd try pushing all the keys to on BUT 3, 9, and 7. I bet that's not right, but it's what I'd try.

[mxysptlk]

12 is easy
it is
4 13 38 87 168 289
isn't it
4-13=9 3*3=9
38-13=25 5*5=25
and so on
just and the next odd numbers square

[DJC]

10. is solvable assuming that that means explaining how it is possible fort D to know. I don't believe that we can infer what the two numbers are. The fact that Q doesn't know, rules out a lot of quotients and means that there are several differences that D might have that would be unique and each implies a different pair of numbers. This is provided we are meant to infer that both numbers are between 2 and 20 - the wording is odd.

[mxysptlk]

if the palindrome one is like this
01:00:10
01:11:10
there would be about 6 an hour and with 24 hours you get 24*6=144sec or 2min 24sec

[Ghost Post]

11. Let's try the obvious other way. Hit every button EXCEPT 3,9,7. Unfortunately there is nothing in this puzzle to let you know if you have solved it.

[Jen Aside]

The palindrome one is for Round 3, not the Porpourri. In either case, you don't count, for instance, 19:00:91, because there are only 60 seconds in a minute.

[Jen Aside]

That's what I said earlier, Psychlogic

[DJC]

On 7, I only get two feasible combinations of numbers. They must multiply to a power of 10 and be distinct. I got 1,2,4,5,10,25 and 1,4,5,10,20,25. If this is all, odds are 1 in 2, but I may have missed some.

On 11, I agree that there is no unique answer. However, assuming the number pattern matches a telephone or standard security keypad, then 1,3,9,7 covers the corners of the largest square. Using the principle that the 'right' answer to such puzzles is one that uses the most information, then this has some appeal.

[DJC]

On 1, it wasn't pretty till I got to the end. Trick is to work with events that aren't conditional on a mass of alternative prior events or the problem blows up. Found it easiest to start at the end - where the only possibilities are the last vote for A or the last vote for B, both with known probabilities, - and work backwards. Still requires an educated guess followed by induction. Anyway, any takers for ((x-y)+1)/(x+1)? Certainly seems right for small numbers. Anyone have a more elegant approach, because mine certainly isn't?

[CrystyB]

5 (Chess) k i plugged this in my CM2k and found that the solution i gave isn't the correct one. But because i was cheating, i can only VERIFY other people's solutions from now on. I won't post what i got.

7 1 2 5 10 25 40 & 1 5 10 20 25 40 DON'T challenge my skills of working with Excel!

10 isn't solvable.

[ralphmerridew]

For the safe combination, maybe the answer is to take the binary representation of 397, and push the buttons that correspond to 1's.

(Note: I think that the answer involves Three hundred ninety-seven, not Three, Nine & Seven; when order is irrelevant, people tend to put numbers in order.)

[Ghost Post]

I get (x-y)/(x+y) for #1.


Consider a grid in which the x-axis represents the number of votes counted. The y-axis represents the current tally, expressed in terms of the difference in the vote count. Thus, the point (1,1) indicates that after 1 vote, A is ahead by 1 vote. The point (12,-2) indicates that after 12 votes, B is ahead by 2 votes. The counting process can be represented as a series of line segments connecting all these points. Call one such series a "path". The problem is then “what fraction of the number of possible paths do not touch the x-axis?”

Now, the total number of possible paths is (x+y)!/(x! y!). Call this R. How many of these paths do not touch the axis?

Note that each path must end at (x+y,x-y) and that all paths that do not reach the x-axis must start at (1,1). The problem becomes “easy” when you recognize that the number of paths that go from (1,1) to (x+y,x-y) and that touch the x-axis ... is exactly the same as the number that go from (1,-1) to (x+y,x-y).

To understand this 1-1 correspondence, consider a path that starts at (1,-1). It must cross the x-axis at some point. To find the corresponding path, simply reflect part of that path about the x-axis. In doing so, reflect only the portion of the path from (1,-1) to the first contact with the x-axis. See the following as an example.

<Unfortunately, I don't know how to insert the picture here. I drew it in Word. Suggestions?>

The trick is to understand that there must be a 1-1 correspondence between these paths.

Now, the number of paths that go from (1,-1) to (x+y,x-y) is (x+y-1)!/(x! (y-1)!). This is precisely the number of paths that go from (1,1) to (x+y,x-y) and that do touch the x-axis. Therefore, the number of paths that go from (1,1) to (1+y,x-y) without touching the x-axis is (x+y-1)!/((x-1)! y!) - (x+y-1)!/(x! (y-1)!). This can be simplified to (x-y)(x+y-1)!/(x! y!). Call this Q.

So, the probability that A is always ahead is Q/R = (x-y)/(x+y).


[This message has been edited by shoover4 (edited 09-21-2001).]

[CrystyB]

ok, then about 397... In base 10 it is prime, so it can't be decomposed to say 17, 23 and 5. In base 2 (10 digits) it is 0110001101. Hope this helps anyone.

[Ghost Post]

Sorry, Jen Aside. didn't see your answer.

[Kady]

For that 397 one, how about pushing in the first 3, then pushing 9 in, then 7, which would leave 1, 2, 3, 8, 9 and 0 pushed in and 4, 5, 6, and 7 out.

It's a stupid idea, but an idea nonetheless.

Kd

[Marvin]

For 8, the logic puzzle, I can only count 4 building types, but 6 hobbies / occupations (gardener, figure skater, ice fisher, jogger, Hapkido Instructor and Xerxes works at the zoo).

[DJC]

ChrystyB, one of us must be reading Q10 differently from the other. There is nothing to say the quotients need to be integers. Suppose I am D and know the difference is 16. Then there are 3 possibilities - (20,4), (19,3), (18,2). I don't know the answer to the number pair on this information alone. Then Q tells me the quotient is not unique across the integer pairs between 2 and 20, largest on smallest. These quotients are 5, 19/3 and 18/2. Then I know that the last two of these options are out - 19/3 cannot be further reduced, and 18/2 cannot be reduced while keeping the denominator in the range of 2 to 20. The first quotient is ambiguous - it could result from (20,4) or (10,2) or (15,3). I can therefore conclude that the quotient is 5 and that the correct pair, matching both quotient and difference, is (20,4). Am I missing something (and my apologies if I am)?

By my reading, the fact that Q doesn't know from the quotient alone does rule out a lot of quotients - 20/2, 19/2, 19/3, 19/4, ...,18/5...In fact, every ratio of numbers in which the numerator is a prime greater than 10; every ratio where the numerator is greater than 10 and has no factors in common with the denominator; or the numerator is greater than 10 and the denominator is 2. The last arises because both numbers must be at least 2, and this allows the above solution.

On the voting problem, I agree that Shoover4 seems to have provided a very neat solution. My earlier attempted solution was to the wrong problem - ie, what is the probability that A is always greater than or equal to B (rather than just greater than)? I will be keen to explore whether Shoover's symmetry argument can be translated to give an alternative and easier resolution of this alternative problem.

[Hagar]

A lottery is a lottery. Can someone tell me why it is important *how* Prof Gamble chose his numbers?

[Ghost Post]

planet_buzz it could be kb->"The Stand"->EH he was uncredited in the tv movie but he was in it.

[CrystyB]

Marvin, working at the zoo doesn't count. It's just an unrequired information that made the statement contain more useless details . Also, i noticed that the names all start with consecutive letters, in groups of a..e, f..j, k..o, p..t and (finally, the NAMES - these should've been the top of the list...) u..y, that's why i asked what building type starts with d...

DJC, you're coincidentally right. I mean, you used THE ONLY POSSIBLE SOLUTION "by your reading". I just posted that above. So i guess you're backing me up with that solution, TY. BUT the quotient noun i usually use means integer quotient, with remainder unimportant... I guess i will wait for mith to clear that up then.

[Ghost Post]

2 I agree with Jen aside

3 Tapestries
Mensa

4 Has to be a motor bike and with a hungry roar possibly a Harley

10 I agree with CrystyB 20,4

[DJC]

Apologies, ChrystyB, I misunderstood your last line re the decimals in the earlier post. I must say that, in quite a few years of working in the field, I have never interpreted 'quotient' as being limited to integers without an explicit restriction of the problem to the set of integers, such as is sometimes done in Number Theory. IQs certainly record non-integer quotients (ie percentages), and the quotient rule in calculus deals with continuous, not integer, functions.

But that is a semantic issue. With your definition there would be no solution; with one allowing non-integers there is just one (I now realise, not have checked exhaustively earlier, because my interpretation of the problem did not require that we determine a unique pair of numbers, but instead only establish that D could infer a unique pair of numbers with the extra information D had that we had not been given - that the quotient was 5). As it turns out, it is possible to solve for a unique solution; you posted it and I hadn't checked for uniqueness.

I am not sure I agree with your suggestion that I was correct 'coincidentally', though I understand what you mean. I was responding only to your earlier argument that I was wrong in saying that the non-uniqueness of the quotient rules out a lot of possibilities - and I tried to show that an explanation for Q and D's statements was possible. I do endorse your solution under the wider interpretation of 'quotient', and expect Mith to endorse it also, and the associated meaning of 'quotient' in the question.

[Top Gun]

planet_buzz - EH was the Hero in The Abyss

ie.

The Rock -> EH -> The Abyss -> MB

[g0ph3r]

Q:11 i noticed in the binary representation of 397 there are ten digits. what if those ten digits correspond to the ten keys on the keypad.... 0 being off and 1 being on, or vice-versa.

[impossibleroot]

g0ph3r, I think you've got it. The lines "...is through the use of 10 buttons, which can be pushed in and out" sounds suspiciously like binary (0=out, 1=in).

In that case, the combo is 2-3-7-8-10, or 1-2-6-7-9, depending on where the 0 or 10 is. (0110001101)

[Ghost Post]

DJC,

I haven't been able to apply the symmetry trick to the problem you suggest (what is the probability that A is never behind?). Please let me know if/when you make any progress on it. That problem is substantially more difficult (I think).

[Marvin]

CrystyB - for 8, do the solutions come out with the same answer for the 'Hapkido Instructor'?
I spent some time trying to figure it out (I had to go to work Sunday to watch some dumb database scripts run - it was either figure this out or gnaw off an ankle out of boredom).
I ended up with a standard logic puzzle grid with a load of ticks and crosses on but no unique solution either. I haven't tried coming up with all possible consistent solutions though.

[This message has been edited by Marvin (edited 09-26-2001).]

[DJC]

Shoover4, I am fairly confident my earlier answer to the 'modified' vote counting problem is correct - proof by induction is relatively straightforward once the pattern is discerned, and the formula works for the first couple. However, like you I have had difficulties with adapting your path analysis when we move away from the strict inequality, and I suspect the symmetry cannot work without a tweak - because you double count those paths that touch the axis. If I get time, will look at whether I can solve exactly for the number of paths that touch, but don't cross, the axis and this should add to your solution, but am not very optimistic (short of using induction, that rather defeats the purpose). Would welcome an earlier insight.

[Ghost Post]

Newbie alert!

Hi all!

Anyway, Q7, lottery. I read this and I thought it was immediately obvious that all the logarithm stuff was a distraction, but obviously no one but Hagar spotted it.

So the solution:

Probability is 1 in n!
-----------
r! * (n-r)!
Where n is 46 and r is 6.
Which gives:
1 in 9,366,819

Adrian.

------------------
--
To a worm, even digging in the hard ground is more relaxing than fishing.

[Marvin]

The way I read the 'base 10 log' stuff was that when you multiply the Prof's numbers together, the product is a power of 10 (1000, 10000, 100000 etc). Ditto for the winning numbers.
There's only four sets of 6 numbers between 1 and 46 that satisfy this. The Prof has one of these combinations, so his odds are 1 in 4.

[The way I got only 4 sets of numbers is by decomposing the required totals (1000, 10000 etc) into their primes. They are all multiples of 2 and 5, with the same number of 2's as 5's.
1000 = (2 * 5) * (2 * 5) * (2 * 5) etc
You then have to choose sets of 6 numbers that are all multiples of 2, 5 or both 2 and 5 so that when you miltiply them togther, you have the same number of 2's as 5's)

[Marvin]

I don't think anyone tried Quailman's 3x3. How about:
HOT: Pink, house, white
BOAT: House, river, ?
SITTING: Duck, baby, ?
The only things I can think of for 'dung' are beetle, spreader and perhaps cow.

[Ghost Post]

the lottery answer is the trick question this time around. Notice that the riddle says that the numbers he picked and the numbers on his ticket have the same properties. The solution is 1 in 46^6

[Ghost Post]

the lottery answer is the trick question this time around. Notice that the riddle says that the numbers he picked and the numbers on his ticket have the same properties. The solution is 1 in 46^6

[Ghost Post]

THe chess puzzle:

What's wrong with playing the queen to the left one, then to the back rank. The king is paralyzed, and the bishop cant cover it. Then to mate, double check the king with the knight, and a discover check from the queen, i.e. N-G6