Math Puzzles (Read the Rules!)

[ChUcK]

I got one for you- How many licks does it take to get to the tootsie-roll center of a tootsie pop? Please include a chart relating solubilities of various sugars and dyes in saliva, average mass of candy vs average mass of the tootsie roll center of a tootsie pop, and coefficients of friction between tongue and candy.

Mr. Owl thinks he's so smart. Smarmy strigidae, thinks he can play by his own rules. We'll show him.

[veganray]

OK, I'm a bit bored. Here's an easy one:
There are 20 quarters sitting on a table, randomly arranged: 10 heads and 10 tails. You are sitting at the table with a blindfold and gloves on. You are able to feel where the quarters are, but are unable to see or feel if they heads or tails.

The task with which you are charged is to make two sets of 10 quarters each. Each set must have exactly the same number of heads and tails as the other set. You may only slide or flip the quarters; you are unable to determine the heads/tails status of any quarter at any time.

What algorithm do you employ to infallibly create two 10-coin sets with the same number of heads and tails in each set?

[pdorries]

seperate the 20 coins into two different groups of 10, take one of the groups and flip all the coins, they should be the same number of heads and tails as the other group

[veganray]

Told y'all it was easy. Even pdorries could get it!

[pdorries]

hey thats not very nice! lol. i will admit I did have to think about it for about 7 minutes.. i almost gave up but then I thought to myself it has to be easy because theres not many rules or combinations.


okay... this one is probably pretty easy too, just takes a little time...

Let me preface this story... This is the coolest disc golf hole in the world! It starts on a hill and you have to throw down a tunnel no bigger than two people wide, you are throwing downhill, and it is over a rushing stream. you have to retee if you don't make it to the green. There is a wooden bridge you walk across to get to the green. The only rules are, to play this disc golf hole, you have to play it with someone else (two at a time), and one of you has to come back to the people up top until everyone has played the hole and putted out (people who lose their discs have to retrieve them with the golden retriever, so every group needs the retriever).

Four people want to play this incredibly amazing, lush, gorgeous hole 18. Veganray takes 1 minute to play the hole because he is exceptionally good and has pinpoint accuracy and distance control, Trey133 (from here on known as Sir Treybroham Lincoln) takes 2 minutes because he usually needs one drive to get adjusted to the setting and then is good to go, pdorries takes 5 minutes because he likes to look at all the scenery as hes walking down the bridge to the green, and CHUCK takes 10 minutes because he ends up emptying his bag off the tee, only to get his only innova disc (a beat up dx teebird) to land inbounds (all the Discrap went in the water).

When a person will cross the bridge, he must be with a person and a person must go back to return to play the hole again (they only have one golden retriever and need to have it available for each group) When crossing, the person with the longest time will be added Example: CHUCK(10 minutes)and Sir Treybroham Lincoln (2 minutes) so 10 minutes will be added. Complete the hole in 17 minutes. Tell who is with who and who will come back to return the golden retriever (it takes the same amount of time to return to the top as it does for that person to play the hole solo, ex. veganray can make it back up top in 1 min)

Hint/clarification: the retriever must return to the top after a group has finished, but anyone who is at the bottom already can return to the top with the retriever at their alotted speed.

[ChUcK]

Oh man, I just spent a half hour writing a sweet narrative answer to the problem until I realized that my solution broke the rules.

Can someone leave the green and meet the group that is playing while they are playing the hole in order to hand off the Gretriever and absorb their return time?

Like, could Trey meet ChUcK and pdorries while they are playing, and the Gretriever and Trey make it back up to the tee before they hole out?

[pdorries]

sorry, it can be done by simply using the rules stated. There's no meeting half way, although that is a clever idea, haha.

Just to make sure the rules are clear, a sample solution of how the process works would be....

vegan and chuck play the hole (10 minutes) vegan goes back (1 minute) pdorries and sir treybroham play the hole (5 minutes) and Sir treybroham goes back (2 minutes) Vegan and sir treybroham lincoln play the hole (2 minutes) for a total time of 20 minutes.

The task is to complete the hole in 17 minutes.

[JohnnyB]

9x - 7i > 3(3x-7u), for i

[veganray]

Awww, you so sweet, Johnny!

[pdorries]

well since its been a while, veganray + trey (2 min), trey goes back (2 min), pdorries + CHUCK (10 min), vegan ray goes back (1 min), veganray + trey (2 min) = 17 min

someone post a new one up!

[Timko]

Oh, you all are so meat.

Separate the red and blue points without using a polynomial greater than 1 or exponential term.

[Parks]

Oh, you all are so meat.

Separate the red and blue points without using a polynomial greater than 1 or exponential term.

Huh?

[StumperTX]

So I thought VeganRay's solution wouldn't work when the Red/Blue distribution was out on the fringes (like 90/10) on the virgin problem, so I built up an excel model of it, and holy crap it actually works for any distribution of Red and Blue hats.

The decision is a hell of a thing to model after the first girl:

=IF($F$5="R",IF(OR(AND(G5=ODD(G5),H6=ODD(H6)),AND(G5<>ODD(G5),H6<>ODD(H6))),"B","R"),IF(OR(AND(G5=ODD(G5),H6=ODD(H6)),AND(G5<>ODD(G5),H6<>ODD(H6))),"R","B"))

I could only get 2 to die (the first two), and that's with a %Red of about 70 or greater.
(Col G Is the # of Blue guessed, H is the Number of Blue ahead, F5 is the first guess)

I may have improved on it a little bit by substituting
=IF(NumBlue=ODD(NumBlue),"R","B")
with
=IF(NumBlue>49,"R","B")
but that's marginal

PM me your email if you'd like the Spreadsheet model to prove it out.

[Timko]

Oh, you all are so meat.

Separate the red and blue points without using a polynomial greater than 1 or exponential term.

seperately the 2 sets with something linear.

Huh?

[Parks]

I seriously don't get what you are asking. And I'm clearly not the only one, given that no one has even attempted a solution.

Can you show us a failed example or something?