[Two boys]

ah, thank you.

[Two boys]

4 hours ago, bonanova said:

@ThunderCloud you're homing in on it, but now you're a little high.

@BMAD It's certainly true that if the FIRST child was a boy born on a tuesday, then it's just the prob that the second child is a boy. But ... the OP does not tell you that. That is, "one is a boy" does not imply "my oldest child is a boy." So your "second" child simply means the "other" child.

Alas, English.... But I am confused. From your interviewing you know for a fact that this family has A boy. So regardless of their birth order all we don't know is the probability that the other child is a boy. And the other child could be a boy or girl, so I still believe it is 1/2.

[Two boys]

On 4/10/2018 at 3:38 AM, bonanova said:

I ask people at random if they have two children and also if one is a boy born on a tuesday. After a long search I finally find someone who answers yes. What is the probability that this person has two boys? Assume an equal chance of giving birth to either sex and an equal chance to giving birth on any day.

If it is given that they already have one boy then... 

I get 1/2. Since the second child could either be a boy or a girl. 

Note: The spoiler buttons seems to be missing on mobile again. 

[Comparing a rectangle to a circle]

Assume we have a circle and a rectangle.  The circumference of the circle is the same as the perimeter of the rectangle and the areas of each shape is also the same.  Write the ratio of the circle's circumference to the diagonal of the rectangle as a function in terms of the width of the rectangle.

[a^b x c^a = abca]

I hope there is a nuanced way of solving this besides brute force.

[Limit of a shrinking function]

2 hours ago, bonanova said:

Which way should we evaluate this?

[ [ [ [ [ x^(x/2) ]^(x/4) ]^(x/8) ]^(x/16) ]^(x/32) ]....

or

x^[ (x/2)^[ (x/4)^[ (x/8)^[ (x/16)^[ (x/32) .... ] ] ] ] ]

The first case

[Limit of a shrinking function]

Find the limit of

x^(x/2)^(x/4)^(x/8)^(x/16)^(x/32)....

(a) as x goes to infinity

(b) as x goes to zero

[Balancing weights]

43 minutes ago, Donald Cartmill said:

4 would be the smallest number.  assign a unit weight of 1 ;2 ;3 ;4; to the 4 weights  place units 1 and 4 on left pan and 2 and 3 on the right pan each weigh 5 and balance

sorry I misunderstood the question in my earlier answer

but if you place 3,4 on one side it would not be balanced.

[I want the butterscotch]

18 minutes ago, plasmid said:

Two more clarification questions

  Hide contents

First, I see no reason to ever use the 75 cent option. If you pay 75 cents to get two pieces at random and you don’t return a piece, then you would have been better off using the 25 cent option twice and spending 50 instead of 75 cents. If you pay 75 cents to get two pieces at random and do return a piece, then you would have only spent 50 cents, but you would have gotten two random pieces and returned a non-butterscotch piece. You would have been better off paying 25 cents twice to get two random pieces without returning a non-butterscotch, because if you do return a non-butterscotch then that just decreases your chances of getting a butterscotch on subsequent rounds. If that analysis is wrong because of an incorrect assumption then let me know.

Second, I think it could make a difference exactly how the 1.50 option works. Specifically, if the machine would have randomly pulled out five non-butterscotch candies, would it drop one of them and pick up a butterscotch? Or would it drop all of them back into the bin and mix all the candies still in the bin and then draw five more random candies? Or would it set the five that it picked aside and draw five more candies that were not in its initial draw before dumping everything back into the bin?

 

To answer your second part: the machine would scan the candies and intentionally pick out a butterscotch and randomly select four candies from the remaining four. So in this case, if there is a butterscotch candy left then you are guaranteed that the first one chosen was a butterscotch. 

[Help Bob find Alice's P(x)$%^]

n is not known

[Help Bob find Alice's P(x)$%^]

Alice and Bob are playing the following game: Alice has a secret polynomial P(x) = a_0 + a_1 x + a_2 x^2 + … + a_n x^n, with non-negative integer coefficients a_0, a_1, …, a_n. At each turn, Bob picks an integer k and Alice tells Bob the value of P(k). Find, as a function of the degree n, the minimum number of turns Bob needs to completely determine Alice’s polynomial P(x).

[Is it a factor?]

Consider the set {1,11,111, …, ((10^2007) – 1)/9}.  At least one of these numbers is divisible by 2007.

Is the same true for 2008 (replacing 10^2007 with 10^2008, of course)?

[Code to the safe]

You have 7 generals and a safe with many locks. You assign the generals keys in such a way that EVERY set of four generals has enough keys between them to open ALL the locks; however, NO set of three generals is able to open ALL the locks. How many locks do you need, and list how many keys does the first general get, the second, … Is there more than one way that works?

[I want the butterscotch]

14 hours ago, Molly Mae said:

Clarification question:

 

  Hide contents

If I put in 75p, can I return a candy I received from a previous purchase for the 25p rebate?

 

Yes, the machine only requires that one of its candies be returned it need not be one purchased with the 75 cents.

[I want the butterscotch]

There is a machine with 20 pieces of candy.  Five of those candies are butterscotch.  If you put in a 25 cents, one candy is provided at random.  If you put in 75 cents, two candies are dropped at random but you may give the machine back one candy in exchange for a 25 cents.  And if you put in $1.50 you receive 5 pieces of candy at random but are guaranteed at least one butterscotch.

How much should I expect to spend to get all of the butterscotch?

[Balancing weights]

On 3/9/2018 at 6:55 PM, plainglazed said:

another seven

  Reveal hidden contents

1,2,3,4,5,6,7

and I think an example of six?

  Hide contents

8,7,6,5,4,2

still working on five...

multiples of your six seem to work too

[Balancing weights]

On 3/10/2018 at 3:17 AM, plainglazed said:

for a set of five...

  Hide contents

...thinking it is not possible.  Label the five integers in increasing order a,b,c,d,e:
e+d has to equal a+b+c and e+c must equal a+b+d but e+d>e+c and a+b+c<a+b+d

 

This contradicts the solution below.  As your a > b > c > d > e > f > g solution does not follow the condition that 

a + b must equal the sum of the rest

On 3/9/2018 at 6:55 PM, plainglazed said:

another seven

  Hide contents

1,2,3,4,5,6,7

and I think an example of six?

  Hide contents

8,7,6,5,4,2

still working on five...

 

[Balancing weights]

A balance and a set of metal weights are given, with no two the same. If any pair of these weights is placed in the left pan of the balance, then it is always possible to counterbalance them with one or several of the remaining weights placed in the right pan. What is the smallest possible number of weights in the set?

[derivative brain teaser]

let a = ln(ln(x))/(ln(x)

let b= x^a

let y = e^b

find dy/dx

[A family of functions]

Find all continuous positive functions that possess the following property:

integral from 0 to 1 of f(x) dx = 1

and integral from 0 to 1 of x*f(x) dx = a

and integral from 0 to 1 of x²*f(x) dx = a²

[Video Game Logic]

I am not disputing either argument, I am curious though on your thoughts that something like the pistol would be better than the shotgun because

2/3's of the time it is equal or better than the shotgun and only 1/3 of the time is it not.

[Video Game Logic]

I was playing a mobile video game based on the fall out series.  In this game, you need to arm your citizens with weapons to protect themselves.  When the citizens use the weapons they will hit their target for an amount of damaged indicated within the range given.  Currently, I have a surplus of weapons and would like to allocate the best weapons to my people.  Place the weapons in order of best to worst:

Knife: 0 - 9 damage

Shotgun: 6 damage

Pistol: 5 - 7 damage

Rail gun: 4 - 7 damage

Rifle: 3 - 7 damage

Grenade: 0, 11 (only 2 possible outcomes)

[you are in an orgy]

14 hours ago, harey said:

So far, so good?
 

  Hide contents

 

Case is M1 or W1 infected.

In brackets:
- empty: none infected on the end of the round
- M1W1:  M1 infects W1 or inversely
- m4W1:  m4 is infected by W1
- ****:  both already infected

Round 1: M1W1 (M1W1) M2W2 (    ) M3W3 (    ) M4W4 (    )
Round 2: M1W2 (M1w2) M2W3 (    ) M3W4 (    ) M4W5 (    )
Round 3: M1W3 (M1w3) M2W4 (    ) M3W5 (    ) M4W1 (m4W1)
Round 4: M1W4 (M1w4) M2W5 (    ) M3W1 (m3W1) M4W2 (****)
Round 5: M1W5 (M1w5) M2W1 (m2W1) M3W2 (****) M4W3 (****)

W1: protection in the 1st round (supposing it is M1 who is infected)
W2: cannot escape
W3: cannot escape
W4: 1st round
W5: last round

At least, did I understand the problem?

 

 

so far, so good

[making obtuse triangles]

Working with lengths that are whole units from 1 - 100, how many obtuse triangles can be formed?

[Circus Bears]

At a circus there was a stunt completed by six bears, 3 brown, and 3 black.  Initially, the 3 brown bears were each standing in a square to the left of a center square and the 3 black bears were located right of center.  The bears either scooted left/right one square or one of the bears jumped over another until each group of bears were on the other group's set of squares.  If the bears needed 15 moves (scoots, jumps) to completely switch sides, how many scoots and how many jumps were used?  Bonus points: what is the largest total amount of bears (assuming equal amounts of brown and black bears) that could complete this task in fifteen moves?

Start: br br br ___ bl bl bl

End: bl bl bl ___ br br br