Challenge Riddles and Puzzles: Extend the following to a valid equation

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The discussion revolves around extending mathematical expressions to create valid equations, with participants sharing various solutions for sequences like 9 9 9 = 6 and 8 8 8 = 6. A secondary topic involves a game played on a circular field where players place coins, prompting questions about the fairness of the game based on the ratio of the field's radius to the coin's radius. Participants express confusion over the concept of fairness in deterministic games that cannot end in a draw, leading to deeper discussions about game theory. Additionally, there are puzzles involving number functions and urns with colored balls, showcasing the group's engagement with logical reasoning and problem-solving. The conversation highlights the blend of mathematical challenges and playful banter among participants.
  • #271
#60
The king moves from position (0,0) to (n,n) where n is 7 for a standard chessboard, and moves with combinations of horizontal, (1,0), vertical, (0,1), and diagonal, (1,1) moves.

For k diagonal moves, the number of horizontal moves needed is (n-k), and the number of vertical moves needed is likewise (n-k). This gives a total of (2n-k) moves.

The total number of permutations with all distinct is (2n-k)!. However, all the horizontal moves look alike, meaning that one must divided by the factorial of their number, (n-k)!. Likewise for vertical moves, (n-k)!, and diagonal moves, k!. Thus giving
$$ N(n,k) = \frac{(2n-k)!}{k! (n-k)! (n-k)!} $$
The total number of permutations is added up over k = 0 to n, giving
$$ N(n) = \sum_{k=0}^n N(n,k) $$
Doing this calculation for n = 7 gives 48639.

I looked for patterns for N(n) as a function of n, without much success. By numerical calculation, I found an asymptotic limit of C0*Cn, for C near 5.8.
 
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  • #272
jbriggs444 said:
Sensible. So one can catalogue the ways of doing it with 0, 1, 2, 3, 4, 5, 6 and 7 diagonal moves:
$$\frac{14!}{7! 7! 0!} + \frac{13!}{6! 6! 1!} + \frac{12!}{5! 5! 2!} + \frac{11!}{4! 4! 3!} + \frac{10!}{3! 3! 4!} + \frac{9!}{2! 2! 5!} + \frac{8!}{1! 1! 6!} + \frac{7!}{0! 0! 7!}$$
$$= 3432 + 12012 + 16632 + 11550 + 4200 + 756 + 56 + 1$$
$$=48639$$

Or one could brute force it with a kind of Pascal's triangle variant
Code: 
     1     1     1     1     1     1     1     1
     1     3     5     7     9    11    13    15
     1     5    13    25    41    61    85   113
     1     7    25    63   129   231   377   575
     1     9    41   129   321   681  1289  2241
     1    11    61   231   681  1683  3653  7183
     1    13    85   377  1289  3653  8989 19825
     1    15   113   575  2241  7183 19825 48639
lpetrich said:
#60
The king moves from position (0,0) to (n,n) where n is 7 for a standard chessboard, and moves with combinations of horizontal, (1,0), vertical, (0,1), and diagonal, (1,1) moves.

For k diagonal moves, the number of horizontal moves needed is (n-k), and the number of vertical moves needed is likewise (n-k). This gives a total of (2n-k) moves.

The total number of permutations with all distinct is (2n-k)!. However, all the horizontal moves look alike, meaning that one must divided by the factorial of their number, (n-k)!. Likewise for vertical moves, (n-k)!, and diagonal moves, k!. Thus giving
$$ N(n,k) = \frac{(2n-k)!}{k! (n-k)! (n-k)!} $$
The total number of permutations is added up over k = 0 to n, giving
$$ N(n) = \sum_{k=0}^n N(n,k) $$
Doing this calculation for n = 7 gives 48639.

I looked for patterns for N(n) as a function of n, without much success. By numerical calculation, I found an asymptotic limit of C0*Cn, for C near 5.8.
Not sure whether I should congratulate you. You just won the position of:

"If there's something to count,
and it don't look good:
Who you going to call?" 👻
 
  • #273
61. What is the formula of the number of (connected) squares on a ##n\times n ## chess board.

D77
 
  • #274
For #61, what counts as connected?
 
  • #275
Path connected. Not a square in each corner which combined is again a square. That was just an insurance.
 
  • #276
8x8 of area 1 square, 7x7 of area 22 etc. $$\sum_{n=1}^8 n^2 = 204 $$not as many as I would have guessed
 
  • #277
61.

For the record: It was asked for the formula for the number of squares on a ##n\times n## chess board, so the answer is
$$
\sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}{6}
$$
 
  • #278
62.

43 Lothers befether the Jolice.
41 Lothers sprungle the Leesand.
49 Lothers delorkle the Gmelt.

17 Lothers befether the Jolice and sprungle the Leesand.
27 Lothers befether the Jolice and delorkle the Gmelt.
21 Lothers sprungle the Leesand and delorkle the Gmelt.

6 Lothers do all three.
8 Lothers do nothing at all.

How many Lothers are there?

D78
 
  • #279
fresh_42 said:
62.

43 Lothers befether the Jolice.
41 Lothers sprungle the Leesand.
49 Lothers delorkle the Gmelt.

17 Lothers befether the Jolice and sprungle the Leesand.
27 Lothers befether the Jolice and delorkle the Gmelt.
21 Lothers sprungle the Leesand and delorkle the Gmelt.

6 Lothers do all three.
8 Lothers do nothing at all.

How many Lothers are there?

D78
Likely there is a clever inclusion/exclusion formula that applies. But just as easy do it exhaustively.

6 do all three.

Given this, then of the 21 that both sprungle and delorkle, 15 do only those two things.
Given this, then of the 27 that both befether and delorkle, 21 do only those two things.
Given this, then of the 17 that both befether and sprungle, 11 do only those two things.

Given the above, then of the 49 that delorkle, 7 only delorkle while 15+21+6 also do something else.
Given the above, then of the 41 that sprungle, 9 only springle while 11+15+6 also do something else.
Given the above, then of the 43 that befether, 5 only befethers while 21+11+6 also do something else.

8 do nothing.

The exhaustive catalogue of Lothers then includes 8 who do nothing, 5+9+7 = 21 who do exactly one thing, 15 + 21 + 11 = 47 who do exactly two things and 6 who do three things.

For a total of 8 + 21 + 47 + 6 = 82 Lothers.
 
  • #280
fresh_42 said:
For the record: It was asked for the formula for the number of squares on a ##n\times n## chess board
Aint'no *&&^% chess boards other than with 8x8 squares :oldlaugh: !
 
  • #281
jbriggs444 said:
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

BvU said:
Aint'no *&&^% chess boards other than with 8x8 squares :oldlaugh: !
Are you sure?
 
  • #282
mfb said:
Are you sure?
How disappointing: this one was missing at the link

71LFn8C2tIL._SX425_.jpg
 
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  • #283
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
 
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  • #284
fresh_42 said:
D78
Infinitely many unless n=0.
x, (x), ((x)), (((x)))...

[Waits for the wet noodle]
 
  • #285
jbriggs444 said:
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.
 
  • #286
fresh_42 said:
... 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.
Google says Catalan numbers
 
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  • #287
jbriggs444 said:
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}}## :smile:

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}}##
 
  • #288
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
 
  • #289
fresh_42 said:
D78
[bungled]
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.
 
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  • #290
What is it that you calculated for G2? Smart means G2 > G1 = ##\frac{11}{36}##.
 
  • #291
fresh_42 said:
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*.
 
  • #292
Other opinions out there?
 
  • #293
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
 
  • #294
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.
 
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  • #295
fresh_42 said:
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.
 
  • #296
mfb said:
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\,.)##
 
  • #297
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
 
  • #298
## 200\pi ## cm
 
  • #299
Nope.
 
  • #300
Ah, 100, of course, darn, o:) I should read better (50 is not the radius)
 

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