Riddles and Puzzles: Extend the following to a valid equation

[Orodruin]

@lpetrich's solution is correct.

I disagree. In particular

Player #1 still has an opportunity for a two-move win, and plays one of the remaining two cards that will make it possible. I select 8 for definiteness.

is not correct. Under your rules this would bring him to 17 for a certain loss or draw and so taking 8 would not be a good choice unless it is to avoid a loss. Also the assertion that 1,2,3,4,5 would sum to 15 is irrelevant because what matters is what comes before, not only what is about to come.

Imagine the cards are arranged as a magic square:

4 9 2
3 5 7
8 1 6

Now it is tic-tac-toe.

I disagree. It is not tic-tac toe. To consider it tic-tac-toe you must first show that the only combination that can win is with three cards, but both 2, 3, 4, and indeed 5 cards (as in 12345) are possible wins under the rules presented.

The dictionary plus Google translate both said 'alternative'.

Wiktionary says:

alternatively
adverb
1. in an alternative way.
-------
alternatingly
adverb
1. In an alternating manner; taking turns.

Edit: Just as an example. The following (given your board configuration) is a win for x:
o x x
x o _
o x _
in your game, but not in tic-tac-toe. To show that the game is equivalent to tic-tac-toe you must show that these situations cannot occur.

Edit 2: The correct argument for why the game will draw is that there is always at most one card that results in victory for your opponent. To avoid a loss (unless you can win), you must select that card. Your opponent will have the same strategy.

 

[fresh_42]

It is tic-tac-toe since I assumed an optimal strategy from both players. Of course you can lose deliberately. You just cannot win.

 

[Orodruin]

It is tic-tac-toe since I assumed an optimal strategy from both players. Of course you can lose deliberately. You just cannot win.

To show this you still need to show that all 4- and 5-combos that win are unattainable. This is not trivial. Yes, you can show that there will be no 3-combo that wins, but you must show that it is impossible to devise a strategy that wins with 4 or 5 cards (which you, per my argument, cannot do) without winning the 3-in-a-row of tic-tac-toe. I honestly think my argument is much simpler. There are also bad tic-tac-toe moves that will not lose you this game, which to me suggests that the strategy applicable to this game is much easier. Also, the first move of the second player can be forced, unlike in tic-tac-toe.

Edit: Regardless, I do not buy the explanation of #138 as it is clearly based on false assumptions (such as taking 8 after taking 9 leaving you a possibility to win).

 

[fresh_42]

Sure, a formal proof requires some work to do. But this is not the thread for a game theoretic problem. The tic-tac-toe argument is easy to see and sufficient in this frame. I mean, nobody did #27 which is even easier, how can we expect a waterproof argumentation for tic-tac-toe. You could as well object, that we haven't shown that this will lead to a draw either.

 

[Orodruin]

The tic-tac-toe argument is easy to see and sufficient in this frame.

Sorry, but I don't see it. I see too many caveats with that argument for it to hold water even informally. (If it was that easy then it should be possible to at least make it plausible to me that it is viable.) It is also certainly not the argument presented in #138 and I cannot see that argument being correct either.

 

[fresh_42]

A player can always chose the tic-tac-toe strategy $S$ of the square in post #140 above and there is nothing the other one can do about it. If you are first to draw, I will choose my card according to $S$. If I start, then you will have to follow $S$, for otherwise I will win. There will be no more than 3 moves, because it will be decided in 3. Only possibility is, that there is a combination of 15 with 3 cards which isn't a row, column or diagonal of the square. But if so, I will follow $S$ and will win since you opened up the possibility of e.g. a row in 3 moves. There is no way you can beat $S$.

P.S.: You were right with alternating. I made a mistake. Likely due to day nighttime.

 

[Orodruin]

A player can always chose the tic-tac-toe strategy $S$ of the square in post #140 above and there is nothing the other one can do about it. If you are first to draw, I will choose my card according to $S$. If I start, then you will have to follow $S$, for otherwise I will win. There will be no more than 3 moves, because it will be decided in 3. Only possibility is, that there is a combination of 15 with 3 cards which isn't a row, column or diagonal of the square. But if so, I will follow $S$ and will win since you opened up the possibility of e.g. a row in 3 moves. There is no way you can beat $S$.

Tic-tac-toe is not necessarily decided in three moves. Also, tic-tac-toe strategy places less constraints on the second player than this game does, for example for the first move. For example, if you pick 9 then I have to pick 6 in this game or you win. In tic-tac-toe I can pick 5. The only optimal response to me picking 5 in this game is picking 6 to win. In tic-tac-toe it is to force a draw, which can be done with several different moves.

Here is an example of a strategy that draws tic-tac-toe but where deviations can be made to win this game:

x = 0, o = 0
4 9 2
3 5 7
8 1 6

x = 5, o = 2
4 9 o
3 x 7
8 1 6

x = 8, o = 9
4 9 o
x x o
8 1 6

x = 14, o = 13
o 9 o
x x o
8 1 x

Here the correct tic-tac-toe move is for x to take 9, but the correct move in this game is to take 1 for the win. The tic-tac-toe game goes on to:

o x o
x x o
8 o x

o x o
x x o
o o x

and a draw.

Of course, the correct block by o in this game is not 4, but 1, which leads to both players busting.

But why not skip all of these complications and just argue based on "If I cannot win, there is at most one card I need to take to avoid that my opponent wins in the next round." Both players can avoid losing by picking the card their opponent would need the next round and so nobody can win the game since player 1 cannot win with the first move. The argument formally goes:

- The first move cannot win the game.
- Given that move n cannot win the game, there is always a move that prevents a win in move n+1.
- It follows by induction that the game cannot be won in any move (and we just need to carry the induction to n = 9).

 

[fresh_42]

Let's move on, although I still have a little hope that someone would solve #27.

29. A column of $10$ trucks has to cross a narrow wooden bridge the length of which is a whole multiple of the length of a truck. To save the bridge from collapsing, never have $8$ trucks or more at the same time on the bridge. To achieve this, the drivers drive over the bridge at a constant speed, with exactly and optimal half the length of a truck in between. From the moment the first truck drives onto the bridge, to the moment the last leaves the bridge, $10$ minutes and $12$ seconds pass. How long does the passage of one truck take?

 

[mfb]

In tic tac toe, if you respond to "9" with "6" you lose because the first player will continue with 2 and then 5 (getting either 9251 or 9258 to win). In the card game you have to react with 6 or you will lose. I don't see any relevant similarity between the games.
4 9 2
3 5 7
8 1 6

Orodruin has the right argument.

Spoiler: bridge

The bridge is 11 truck lengths long. The critical moment is "truck|gap (truck gap)⁷|truck" where | is the bridge border and seven trucks are on the bridge: The first truck must fully clear the bridge by the time truck 9 starts entering it.
The convoy is 10+9/2=14.5 trucks long. During the given time span it moves by its own length plus the length of the bridge, or 25.5 truck lengths. 612 seconds divided by 25.5 is 24 s. Each truck drives its own length in 24 s.
A single truck will be partially on the bridge while driving 11+1=12 truck lengths, or 12*24 s = 4 minutes 48 seconds.

 

[pbuk]

P.S.: You were right with alternating. I made a mistake. Likely due to day nighttime.

The best word for this use in British English is alternately.

 

[fresh_42]

30. Five old ladies are collecting buttons of all kind. Together they have 2000 pieces.
If we add Elise's buttons to Adele's, subtract Elise's buttons from Berta's, multiply Elise's buttons with Charlotte's, divide Doris' buttons by Elise's, we will always get the same number. Adele is happy that she soon has 100 buttons. How many are missing?

Correction: There are 2000 buttons without counting Elise's.

 

[pbuk]

I mean, nobody did #27 which is even easier

I'm not sure #27 is as easy as you think it is: it is fairly easy to convince yourself that 9 initial cells are not sufficient, but turning this conviction into a proof is another thing!

Spoiler: But i'll try...

1. Mold can spread into a new cell if adjacent cells in the same row and column are moldy. If there are only 9 initial moldy cells, this way mold can only spread into 9 rows and columns leaving 19 cells free.

2. Mold can also spread into a new cell if the two adjacent cells in the same row are moldy. This way a new column can be populated. But if there are initially 2 cells in the same row, then by the argument in 1. mold can only spread into 8 rows, so we have 'bought' another column at the cost of a row.

3. The argument in 2 applies with rows and columns interchanged.

Convinced? No, nor am I.

 

[fresh_42]

I'm not sure #27 is as easy as you think it is: it is fairly easy to convince yourself that 9 initial cells are not sufficient, but turning this conviction into a proof is another thing!

Spoiler: But i'll try...

1. Mold can spread into a new cell if adjacent cells in the same row and column are moldy. If there are only 9 initial moldy cells, this way mold can only spread into 9 rows and columns leaving 19 cells free.

2. Mold can also spread into a new cell if the two adjacent cells in the same row are moldy. This way a new column can be populated. But if there are initially 2 cells in the same row, then by the argument in 1. mold can only spread into 8 rows, so we have 'bought' another column at the cost of a row.

3. The argument in 2 applies with rows and columns interchanged.

Convinced? No, nor am I.

I'm a bit confused, to be honest. It can spread into an empty cell if at least two vertical or horizontal adjacent cells (of this new one) are already occupied.

The easiest way is indeed to count the perimeters.

 

[pbuk]

30. Together they have 2000 pieces. ... How many are missing?

Does this mean that they actually have some number less than 2,000?

Spoiler

Is this number 1,961 i.e. 39 are missing?

 

[fresh_42]

Does this mean that they actually have some number less than 2,000?

Spoiler

Is this number 1,961 i.e. 39 are missing?

I'm afraid it isn't solvable unless we observe that it has to be more than 2,000 buttons.
I overlooked a tiny condition. It are 2,000 buttons without Elise's.

Sorry, I definitely will have to post new questions at daytime.

 

[pbuk]

I'm afraid it isn't solvable unless we observe that it has to be more than 2,000 buttons.

For 1,961 buttons I had the solution A = 85, B = 119, C = 6, D = 1734, E = 17 and there are other solutions with fewer than 2,000. But with the new condition it is too easy

Spoiler

although Adele is now deluding herself with her proximity to 100!

 

[fresh_42]

For 1,961 buttons I had the solution A = 85, B = 119, C = 6, D = 1734, E = 17 and there are other solutions with fewer than 2,000. But with the new condition it is too easy

Spoiler

although Adele is now deluding herself with her proximity to 100!

Nice solution, but I don't get your spoiler.
We have $A+B+C+D=2000$ and $A+E=B-E=CE=D/E$ but I don't see why it's easy.

 

[pbuk]

but I don't see why it's easy.

Oh it is only easy for me because I had already tabulated the solution in my attempt to find a solution to the original problem - unless I am wrong of course!

Spoiler: #30

$A = 76; B = 114; C = 5; D = 1805; E = 19$

 

[fresh_42]

31. A famous mathematician (died $1871$) occasionally bragged that he was exactly $x$ years old in year $x^2$. Another one (possibly not a real person) said in $1925$ that he was exactly $a^2 + b^2$ years old in the year $a^4 + b^4,$ that he was exactly $2c$ years old in $2c^2$ and that he was exactly $3d$ years old in $3d^4$.

What are $x,a,b,c,d$ and do you know who the mathematician was?
Hint: The phrase is especially funny in this case!

 

[lpetrich]

I'll take on the first part of #31.

Spoiler

If that mathematician was x years old in year x², then his birth year was x(x-1). He must have been at least around 10 when he died, so he could have some mathematical ability, and at most around 122, the human longevity record. Since he died in 1871, that means 1749 to 1861. The only integer x that fits is x = 43, with x² = 1849. That makes him born in 1806, and age 65 when he died.

That mathematician was Augustus de Morgan.

 

[lpetrich]

Now the second part of #31.

Spoiler

Since he must have been between 10 and 122 years old in 1925, his birth year must have been between 1803 and 1915.

The d equation means that his birth year must have been 3d⁴-3d. The only birth year in that range for positive-integer d is 1860, for d = 5. That means that he was 15 years old in 1875.

Turning to the c equation, the only positive-integer value of c is c = 31, making him 62 years old in 1922.

Turning to the a and b equation, the birth year is B(a) + B(b) where B(x) = x⁴ - x². This function is monotonically increasing for x > 1, and B(1) = 0, B(2) = 12, B(3) = 72, B(4) = 240, B(5) = 600, B(6) = 1260, and B(7) = 2352. This means that both a and b are at most 6, and that at least one of a and b is at least 5. Considering values 5 and 6, 1860 - B(5) = 1260, and 1860 - B(6) = 600. This means that a and b are 5 and 6.

 

[lpetrich]

Now #30.

Spoiler

For Adele having a buttons, Berta having b buttons, Charlotte having c buttons, Doris having d buttons, and Elise having e buttons, the conditions are:

• a, b, c, d, and e are nonnegative integers
• a + b + c + d = 2000
• a < 100
• a + e = b - e = c*e = d/e = w

I was able to solve these equations using Mathematica's Reduce function. I found a single solution: a=76, b=114, c=5, d=1805 e=19, with Adele needing 24 more buttons.

 

[fresh_42]

32. Two friends are bored playing backgammon and decided to take the dice and roll dice. One takes the four normal dice with numbers $1,2,3,4,5,6$, the other one takes the die with numbers $2,4,8,16,32,64$. Who will win more often?

 

[Orodruin]

What is the win condition? Higher total in a roll?

 

[lpetrich]

I'll try to avoid brute force with #30.

Spoiler

From the fourth condition, a = w - e, b = w + e, c = w/e, d = w*e.

Let a + b + c + d = n. Then n = 2w + w/e + w*e = w*(e+1)²/e.

Since e and e+1 are relatively prime, w must be divisible by e, giving w = v*e. Thus,

a = (v-1)*e, b = (v+1)*e, c = v, d = v*e², n = v*(e+1)²

Here, n = 2000 = 2⁴*5³. This gives possible values for e+1 of +-1, +- 2, +-4, +-5, +-10, and +-20. Since e must be positive, that gives e = 1, 3, 4, 9, and 19, with corresponding values of v of 500, 125, 80, 20, and 5, and values of a of 499, 372, 316, 171, and 76. Of these, only the last one is less than 100.

 

[lpetrich]

I'll take on #32.

Spoiler

Let the first dice have probability distribution p(i) for roll-value i, and the second die have probability distribution q(j) for roll-value j. The total probability that the second die will have a higher number is
$$ \sum_{i<j} p(i) q(j) $$
For the four 1-to-6 dice, one can first find the probabilities for two dice, and then for four, by twice repeating the two-distribution self-convolution:
$$ p'(i) = \sum_j p(j) p(i-j) $$
This can be done with a generating-function approach, and it takes only a few lines of Mathematica code to do that.

I find these probabilities:

• First dice higher: 4127/7776 ~ 0.530736
• Equal: 161/7776 ~ 0.0207047
• Second dice higher: 109/243 ~ 0.44856

So the first player will more likely win.

 

[fresh_42]

What is the win condition? Higher total in a roll?

Yes.

 

[fresh_42]

33. Build with $\{\,0,1,2,4,+,-,\cdot,:\,\}$ all integers $1,2,\ldots,25$.

 

[mfb]

Do we have to use all numbers once, or at most once?
I'll start with 1-9:

Spoiler

4-2-1-0=1
4-2*1-0=2
4-2+1-0=3
4-2*1*0=4
4+2-1-0=5
4+2+1*0=6
4+2+1+0=7
4*2+1*0=8
4*2+1+0=9

 

[fresh_42]

Do we have to use all numbers once, or at most once?
I'll start with 1-9:

Spoiler

4-2-1-0=1
4-2*1-0=2
4-2+1-0=3
4-2*1*0=4
4+2-1-0=5
4+2+1*0=6
4+2+1+0=7
4*2+1*0=8
4*2+1+0=9

The solution I have uses all numbers exactly once, so this would be a nice to have constraint, yes.

 

[mfb]

Is exponentiation allowed? It doesn't have an extra symbol...
A few towards the end, assuming concatenation is okay:

Spoiler

20+4*1=19 = 21-4/2
20 is easy to make if we don't use 1 and 4, or if we can use exponents (20*1⁴)
21+4*0=21 = 40/2+1
22?
24-1+0=23 = 21+4/2
24+1*0=24 = 20+4*1
24+1+0=25 = 21+4+0

 

[fresh_42]

22 goes with exponentiation. I don't understand

20+4*1=19

but 19 can easily be done with all four digits.

Now we need: 10-19 and 22.

 

[mfb]

I don't understand 20+4*1=19 either. Copy&paste error, probably. The right side is wrong, too.

20-1⁴ = 19.
21+4⁰ = 22

Spoiler

10 is trivial if we don't use all numbers, and it is easy with square roots, but I don't find a proper solution for now.
12-4⁰=11
10+4-2 = 12 = 4*(2+1+0) = 12*4⁰
14-2⁰=13
14+0*2=14
2⁴-1+0 = 15 = 20-4-1
2⁴+1*0 = 16 = 10+4+2
2⁴+1+0 = 17 = 21-4+0
10+4*2 = 18
20-1⁴=19 from above

10 is missing.

Outside the range but why not:
2⁴+10 = 26

 

[fresh_42]

10 is missing.

Think of the prime factorization.

And now again one of the kind you do not like: What is the next number?
34.
The author of the first Russian mathematical book was born.
The oldest German wine bill dates to this year.
The Knights Templar was dissolved.
The first Scottish university was founded.
The Viri Mathematici, one of the first books about history of science was published.
Kepler's mother was arrested for suspected witchcraft.
The second most famous English mathematician of his time after Newton dies.
The Connecticut Asylum for the Education and Instruction of Deaf and Dumb Persons was founded.
Max Planck wins the Nobel Prize in physics.
This is the year we are looking for.

 

[mfb]

Think of the prime factorization.

Oh... how could I miss something so simple.
(4+1)*2+0=10
Needs brackets, however, unlike every other number.