Riddles and Puzzles: Extend the following to a valid equation

[mfb]

Likely there is a clever inclusion/exclusion formula that applies.

People who do nothing + sum of the people for one activity - sum of the people for two activities + people for three activities
8+(49+41+43)-(21+27+17)+6 = 82

Aint'no *&&^% chess boards other than with 8x8 squares !

Are you sure?

 

[fresh_42]

Are you sure?

How disappointing: this one was missing at the link

 

[fresh_42]

63. On how many ways can parentheses be set in a product of $n$ numbers, such that always at most two numbers are multiplied?

D78

 

[jbriggs444]

D78

Infinitely many unless n=0.
x, (x), ((x)), (((x)))...

[Waits for the wet noodle]

 

[fresh_42]

Infinitely many unless n=0.
x, (x), ((x)), (((x)))...

[Waits for the wet noodle]

... oh, those mathematicians ... you almost never can use words ... I start to understand what Bourbaki has driven.

Corrected.

 

[jbriggs444]

... oh, those mathematicians ... you almost never can use words ... I start to understand what Bourbaki has driven.

So, rephrasing, you want to know how many (finite) binary trees there are with n leaf nodes with the understanding that each node in the tree is either a leaf node with zero children or an root/intermediate node with exactly two sub-trees, each of which is required to have at least one leaf node.

Let t(n) be the number of such trees with n leaf nodes.

t(0) = 1 -- the empty tree (which can never be a sub-tree hanging off of an intermediate node)
t(1) = 1 -- the tree with one leaf.
t(n) = $\sum_{i=1}^{n-1} t(i) t(n-i)$ -- adding up for the possible left hand and right hand sub-trees.

Unfortunately, that's as far as I can drive toward a solution

And the above is incomplete -- it applies to an ordered list of numbers parenthesized in place. For an arbitrary set of n unique numbers, one would need to multiply by n factorial.

Spoiler

Google says Catalan numbers

 

[fresh_42]

So, rephrasing, you want to know how many (finite) binary trees there are with n leaf nodes with the understanding that each node in the tree is either a leaf node with zero children or an root/intermediate node with exactly two sub-trees, each of which is required to have at least one leaf node.

Let t(n) be the number of such trees with n leaf nodes.

t(0) = 1 -- the empty tree (which can never be a sub-tree hanging off of an intermediate node)
t(1) = 1 -- the tree with one leaf.
t(n) = $\sum_{i=1}^{n-1} t(i) t(n-i)$ -- adding up for the possible left hand and right hand sub-trees.

Unfortunately, that's as far as I can drive toward a solution

And the above is incomplete -- it applies to an ordered list of numbers parenthesized in place. For an arbitrary set of n unique numbers, one would need to multiply by n factorial.

If the numbers are not unique then the problem is not well-posed since the answer depends on the number and multiplicity of the duplicates.

Well, Leonhard said it is $\displaystyle{ \prod_{k=1}^{n-1}\dfrac{4k-2}{k+1}}$

but Catalan defined it as $\displaystyle{t_n=C_{n-1}=\dfrac{1}{n}\binom{2(n-1)}{(n-1)}}$ or better $\displaystyle{t(n+1)=C_n=\binom{2n}{n}-\binom{2n}{n+1}}$

 

[fresh_42]

64. "Yahtzee" is a game of 5 dice, in which certain combinations must be achieved. For this you have 3 rolls free and can choose with at each roll, which dice you leave and with which you want to continue to roll.

A "long street" is one of the following two combinations:
1 2 3 4 5 or 2 3 4 5 6
I only had to roll this combination in a game and got in the first roll:
2 3 3 4 6
That's a hopeful start. But how to continue playing? Take one of the two threes and try to throw the missing 5 with the two remaining throws? The chance to win I call G1.

I had the vague feeling that it could be smart to put the 6 in the dice cup and then fill in the remaining formation 2 3 4 with two dice. After all, the combinations 1 + 5 and 5 + 6 can provide a "long street". The chance here is called G2.

Since it must go fast in the game, I opted for the second variant and had bad luck. Afterwards I wanted to know more about it and started to calculate.

Was I smart or not?

D78

 

[jbriggs444]

D78

[bungled]

Spoiler

Smart.

Original strategy:

With only the single die in the cup there are two consecutive chances to roll a winning 5. Probability of success is 11 in 36.

Chosen alternate strategy:

With two dice in the cup the result of the first roll can be an outright win with a 1+5 (4 of 36) or a 5+6 (4 of 36). [The possibilities are non-intersecting, so the probabilities add]. That's a success rate of 8 in 36 immediately.

The result of the second roll (if needed) can be a success at a rate of 8 in 36. Of course, the second roll attempt is conditional on a first roll failure. So it only adds $\frac{28}{36}$ of 8 in 36. Still, that's better than the 3 in 36 you needed to make it a smart play.

We did not even need to examine the possibility of rolling a 1, 5 or a 6 on the first roll.

 

[fresh_42]

What is it that you calculated for G2? Smart means G2 > G1 = $\frac{11}{36}$.

 

[jbriggs444]

What is it that you calculated for G2? Smart means G2 > G1 = $\frac{11}{36}$.

Question does not ask for G2. It only asks whether G2 > G1. I answered yes. Edit: And bungled it. *sigh*.

 

[fresh_42]

Other opinions out there?

 

[fresh_42]

65. Two people play the following game:

On the blackboard is the number $2.$
Who is to move, replaces the number $n$ on the blackboard by a number $n + d$, where $d$ is an arbitrary divisor of $n$ with $1 \leq d <n.$
The one who first has to write a number greater than $1991$ on the blackboard will have lost.

Who wins the game assuming an optimal strategy, the first player or his opponent?

D80

 

[mfb]

Going back to the dice:
- There is a 2/36 chance to roll 1,5 and a 2/36 chance to roll 5,6, for a 4/36 chance to win with the second roll.
- If we don't win then there is a 7/36 chance of 5,x where x is not 1 or 6. We keep the 5 and reroll the last die, for a 1/3 chance to win (with either 1 or 6) -> 7/36*1/3 chance to win
- If we do not get a 5 there is a 9/36 chance to get a 1 and a 7/36 chance to get a 6 but no 1, combined 16/36. In both cases we reroll one die for a 1/6 chance with the third roll.
- If we don't get 1, 5 or 6, which means just 2,3,4 (9/36 chance) then we reroll both and have a 4/36 chance with the second roll.
All cases are mutually exclusive. Sum: 4/36 + 7/36*1/3+16/36*1/6+9/36*4/36 = 5/18 = 10/36
Rerolling 3 and 6 is a little bit worse than rerolling 3 alone.

 

[mfb]

Who wins the game assuming an optimal strategy, the first player or his opponent?

This is a nice puzzle. You excluded n from being added, therefore the first player must add 1. 3 is a prime again, the second players adds 1 and we get 4. This first player now can add 1 or 2, reaching 5 or 6. If the first player goes to 5 then the second player must go to 6.

If 6 is a winning position, the first player will play 4->5 and win.
If 6 is a losing position, the first player will play 4->6 and win.

We don't even have to consider larger numbers. The first player has a winning strategy, although I have no idea how that strategy looks like.

 

[fresh_42]

We don't even have to consider larger numbers. The first player has a winning strategy, although I have no idea how that strategy looks like.

The winner is the one who finds a number $n$ with $1991-n \,|\, n$, because then he can write directly $1991$ and the opponent must choose a larger number.

For odd $n$, $1991-n$ is even, so $n$ can not divide it. The first player can now ensure that his opponent always gets to deal with odd numbers less than $1991$. (His opponent always leaves him an even number less than $1991$, since the divisors of his odd number are all odd and the sum is even, so he can always choose at least $1\,.)$

 

[fresh_42]

66. To move very heavy loads you sometimes use rolls under the load. Such a roll has a circumference of $50\, \text{cm}$. How far has the load been moved when the rolls have turned once?

D80

 

[BvU]

Spoiler

$200\pi$ cm

 

[fresh_42]

Nope.

 

[BvU]

Ah, 100, of course, darn, I should read better (50 is not the radius)

 

[fresh_42]

67. "When you sit in the tram in the afternoon," says one man, "you get the impression that there are three times as many women as men." His neighbor quickly scanned the number of passengers and said, "That's true, at least for them Tram."

Now the tram stopped, and four times as many women got off as men got in. "Now the relationship is a bit more bearable. There are only twice as many women as men in here," the man said to his neighbor.

At the next stop there was only one woman. Since no one was about to get out, the man said to his neighbor, "Come on, we're going out of here, then the old ratio of 1: 3 is restored."

How many people continued on the train after the men left and the lady got in?

D80

 

[mfb]

Interesting general strategy with the divisor puzzle. I expected this to be one of the puzzles where a general solution can be complicated so after I found a proof that there is a strategy (but not the strategy) I didn't look much further.

 

[lpetrich]

Spoiler

I will express the number of passengers as (number of men, number of women).

Initially, there are $(m_1, 3m_1)$ passengers. At the first stop, $(m_2,-4m_2)$ arrive, since departure is negative arrival. This leaves $(m_3, 2m_3)$.

This part has the solution $m_1 = 6n, m_2 = n, m_3 = 7n$, giving $(6n,18n)$, $(n,-4n)$, $(7n,14n)$.

At the second stop, the arrivals and departures are $(-2,1)$, and the result is $m_4, 3m_4$. Solving gives $n = 1, m_4 = 5$.

Initially, the tram has (6,18). At the first stop, (1,-4) arrive, giving (7,14). At the second stop, (-2,1) arrive, giving (5,15). Thus, a total of 20 people continuing on the tram after its second stop.

 

[fresh_42]

68. A pond owner has an unknown amount of fish in his pond. Since he wants to count them now, he catches 30 fish, marks them with red paint and throws them back into the pond. The next day he catches 40 fish. Two of these fish have a red mark.

How many fish are approximately in the pond?

D82

 

[lpetrich]

#68

Spoiler

This is a classic method of wildlife census-taking.

Let us say that there are $N$ fish in the pond. After the first fish catching, there are now $N_m$ marked fish in it. The second time around, the pond owner caught $N_2$ fish, of which $N_{m2}$ are marked. The probability of a fish being marked is $p_m = N_m/N$, and the most likely number of marked fish caught the second time around is $N_{m2} = N_2 p_m = \frac{N_2 N_m}{N}$. This gives us $N = \frac{N_2 N_m}{N_{m2}}$.

Working out the numbers, $N = \frac{40 \cdot 30}{2} = 600$. So the pond owner has 600 fish.

 

[mfb]

If you do this one fish at a time it has an interesting relation to the birthday problem. Consider a pond with 365 fish in it, fish one at a time at random, mark it. When does the first marked fish appear? This is equivalent to drawing random people from a large population, marking their birthdays, and looking for the first duplicate. The answer is well-known: 23 for a ~50% chance[/url]. If we find the first duplicate after 23 fish then we can use 365 as estimate for the total number of fish. It won't be a good estimate with just a single match, of course, but if we look for the second, third, ... we can get better estimates. With the same number of fish you have more information than the two-batch approach.

 

[BvU]

plus or minus how many ? What's the proper standard deviation estimate ? (in #305)

 

[fresh_42]

plus or minus how many ? What's the proper standard deviation estimate ? (in #305)

Well, this is not the thread to ask for $P(590 \leq X \leq 610) \geq .95$.

 

[fresh_42]

69. What time is it if so many minutes are missing until 9 o'clock, as 40 minutes ago 3 times as many minutes after 6 o'clock had elapsed?

D82

 

[BvU]

Spoiler

8:25
from:180 - x = 3x + 40 => x = 35

 

[fresh_42]

This is an experiment with a self made puzzle. So try not to complain too much.

70. I am the product of two odd primes, and can be written only using $2,3$ and at most two $1$. This representation is asked for. My primes are two out of the first three consecutive prime sequence members of the first (minimal first element) sequence with this property, defined as: the ratio of the predecessors in $\mathbb{N}$ of two consecutive sequence elements is $2$. Btw., for my primes $p$ we have $\left( \dfrac{3}{p} \right) \neq -1\,.$

D82

 

[fresh_42]

Too complicated?

 

[lpetrich]

I have a partial solution:

Spoiler

For decimal digits (0 to 2) of 1, (1) of 2, and (1) of 3 and divisibility by exactly two odd primes, I find three numbers:

• 123 = 3*41
• 213 = 3*71
• 321 = 3*107

But beyond this, I am totally stumped.

 

[fresh_42]

You could solve the recursion and test which values $a_0$ lead to three consecutive primes (in the first four sequence elements.) I guess I should have added this, for otherwise there are too many sequences possible.
Excel or a similar program should do the job.

 

[mfb]

Too confusing.

the ratio of the predecessors in N of two consecutive sequence elements is 2

If p,q are elements then (p-1)/(q-1)=2 or equivalently p=2q-1, okay.

My primes are two out of the first three consecutive prime sequence members of the first (minimal first element) sequence with this property

We look for the sequence with the minimal first element. 2->3->5 is the first set of primes that satisfy the condition above, so it should be the first sequence. As 9 is not prime the sequence ends at 5. 2*3, 2*5, 3*5 don't satisfy the condition on the digits of the product. What am I missing?

$\left( \dfrac{3}{p} \right) \neq -1\,.$

Clearly this is not just a regular fraction. I think I have seen that notation in the context of divisors before but don't remember what it was. It would help to clarify the notation (searching for the usual keywords is completely pointless here). I also don't know why this would be relevant.