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

[fresh_42]

Moving the goal posts eh ... ?

I know, my bad. But 15+6 is too easy.

 

[DavidSnider]

19. Construct $21$ with symbols from $\{\,1,5,6,7,*,/,+,-,(,)\,\}$ but use the digits at most once.

Spoiler: spoiler

6 / (1 - (5/7))

 

[fresh_42]

20. The king gives the convicted one last chance to save his life. The prisoner receives 50 white and 50 black balls, which he may arbitrarily distribute to two identical-looking vessels. The next day he has to randomly pick a vessel and draw a ball out of it. He will be executed at black, pardoned for white.

How must the prisoner distribute the balls so that his chances to survive are as high as possible?

 

[pbuk]

15. What is the smallest prime which includes all ten digits exactly once?

There are of course some 'cheating' answers e.g. 1,023,457,986₂₃ = 1,808,433,654,721₁₀ which is prime (probably not the smallest such prime though).

Or to take the radix cheat further, how about 2_134567890?

 

[mfb]

How must the prisoner distribute the balls so that his chances to survive are as high as possible?

Spoiler

(1 white) and (49 white 50 black).
Proof that it is optimal: More than 50% in a vessel requires more white than black there which forces less than 50% in the other one. 49/99 is the closest you can get to 50% winning chance if you have fewer white than black balls, and 100% winning chance is clearly optimal for the other vessel. This strategy is optimal if one vessel has fewer than 50% winning chance. This strategy also beats 50% in both, therefore it must be optimal overall.

 

[fresh_42]

Correct, with an overall chance of almost $3/4$.

21. Which are the numbers (as symbols) $n_0,\ldots ,n_9$ such that the following statement is true:
'This sentence contains $n_0$ times the $0$, $n_1$ times the $1$, $n_2$ times the $2, \ldots$ and $n_9$ times the $9$'?

(And don't force me to phrase it without backdoors, you know what I mean.)

 

[DavidSnider]

Correct, with an overall chance of almost $3/4$.

21. Which are the numbers (as symbols) $n_0,\ldots ,n_9$ such that the following statement is true:
'This sentence contains $n_0$ times the $0$, $n_1$ times the $1$, $n_2$ times the $2, \ldots$ and $n_9$ times the $9$'?

(And don't force me to phrase it without backdoors, you know what I mean.)

I don't think I understand what this is asking. "This sentence contains n0 times the 0". "The" 0?

 

[fresh_42]

'This sentence contains 1 times the 0 and 2 times 1.' is true. The question is: how do we have to chose the $n_i$ that the sentence with all digits is true.

 

[DavidSnider]

'This sentence contains 1 times the 0 and 2 times 1.' is true. The question is: how do we have to chose the $n_i$ that the sentence with all digits is true.

Spoiler: spoiler

11,10,9,8,7,6,5,4,3,2 -> 0, 10, 18, 24, 28, 30, 30, 28, 24, 18

 

[fresh_42]

Spoiler: spoiler

11,10,9,8,7,6,5,4,3,2 -> 0, 10, 18, 24, 28, 30, 30, 28, 24, 18

What do you mean by this? We need a vector $\vec{n} \in \mathbb{Z}^{10}$ such that

'This sentence contains $n_0$ times $0$, $n_1$ times $1$, $n_2$ times $2$ … and $n_9$ times $9$.'

is true. This implies that $n_i \leq 11$ is an upper bound. But actually no component is greater than nine, so $\vec{n} \in \{\,1,2,\ldots,9\,\}^{10}$.

 

[mfb]

Spoiler: spoiler

11,10,9,8,7,6,5,4,3,2 -> 0, 10, 18, 24, 28, 30, 30, 28, 24, 18

That would need e.g. 11 zeros in the sentence "This sentence has 11 times '0', 10 times '1', ...", but the sentence with your numbers plugged in doesn't have that many (it just has 2).
Maybe it is clearer if we add "digit"?

"This sentence contains n₀ times the digit '0', n₁ times the digit '1', ..."

 

[DavidSnider]

That would need e.g. 11 zeros in the sentence "This sentence has 11 times '0', 10 times '1', ...", but the sentence with your numbers plugged in doesn't have that many (it just has 2).
Maybe it is clearer if we add "digit"?

"This sentence contains n₀ times the digit '0', n₁ times the digit '1', ..."

Ahh I see. "This sentence contains n0 occurances of '0', n1 occurances of '1', "

 

[DavidSnider]

Correct, with an overall chance of almost $3/4$.

21. Which are the numbers (as symbols) $n_0,\ldots ,n_9$ such that the following statement is true:
'This sentence contains $n_0$ times the $0$, $n_1$ times the $1$, $n_2$ times the $2, \ldots$ and $n_9$ times the $9$'?

(And don't force me to phrase it without backdoors, you know what I mean.)

Spoiler: spoiler

1732111211
'This sentence contains
1 times the '0'
7 times the '1'
3 times the '2'
2 times the '3',
1 times the '4',
1 times the '5',
1 times the '6',
2 times the '7',
1 times the '8',
1 times the '9'

 

[fresh_42]

22. There are enough beads in three colors to make bracelets with 6 pearls each. Not all three colors need to be used, i.e. there are e.g. also bracelets with six pearls of the same color. Such a bracelet has no beginning and no end. How many different bracelets are there?

 

[lpetrich]

I solved it with brute force with Mathematica:

Spoiler

There are 92 distinct necklaces to within rotation and reflection symmetry. Without including those symmetries, there are 729.

I will attempt to work it out by hand in my next post.

 

[lpetrich]

Here is my more explicit solution:

Spoiler

For one kind of bead, we get solutions of form rrrrrr, gggggg, bbbbb: 3 solutions.

For two kinds of bead, we get solutions like 5 r's 1 g and similar for gr, rb, br, gb, bg, 4r's 2g's, and 3r's 3g's and similar for rb, gb.

rrrrrg - 1 - 6
rrrrgg rrrgrg rrgrrg - 3 - 18
rrrggg rrgrgg rgrgrg - 3 - 9
For 2 different kinds of beads, 33, for at most 2, 36.

rrrrgb rrrgrb rrgrrb - 3 - 9
rrrggb rrgrgb rgrrgb grrrgb rrggrb rgrgrb - 6 - 36
rrggbb - 1 - 1
rgrgbb rggrbb - 2 - 6
rgrbgb - 1 - 3
rgbrgb - 1 - 1
For 3 different kinds of beads, 56, for at most 3, 92.

I could try to derive some general formula.

 

[mfb]

Alternative counting method:

Spoiler

Notation: XYZ are colors where multiple letters can be the same color, xyz are distinct (but unspecified) colors.

Group by symmetry.

• 60 degree symmetry: 3 bracelets (the uni-color ones)
• 120 degree symmetry but no 180 degree symmetry: 3 (xyxyxy)
• 180 degree symmetry: 3*3*3 options for three beads in fixed order. 3 options are unicolor and counted before, leaving 24. As XYZ=YZX=ZXY we divide by 3 to get 8. In addition xyz=zyx from mirror symmetry, which only matters if all three colors are used, reducing the number by 1. Cross-check: Truly different options are xxy (3*2 options for the colors) and xyz (1 option). In total: 7
• Mirror symmetry across two beads but no rotation symmetry: We are free to choose 4 beads (2 on the symmetry axis, two outside , 3⁴=81 options. Again 3 are unicolor, 6 have 120 degree symmetry (xyxy) and 6 have 180 degree symmetry (xyyx), leaving 66. We double-counted (XYZA=AZYX), leaving 33 truly different options.
• Mirror symmetry through edges: We are free to choose 3 beads (27 options) and subtract unicolor (3) and 180 degree symmetry (6, xyx), leaving 18. By avoiding the 180 degree symmetry we also avoid a double mirror symmetry. Again we double-counted so we get 9. Cross check: 3 from xyz-zyx, 6 from xxy-yxx. Fits -> 9.
• No symmetry. If we ignore symmetries there are 3⁶=729 options. In general 12 options lead to the same bracelet so we have to divide - but we have to take care of the symmetric cases separately. We subtract 3 for the unicolor bracelets (just one orientation), 3*2=6 for the 120 degree symmetry (two orientations), 3³-3=24 for the 180 degree symmetry (but not unicolor), 33*6=198 for the first mirror symmetry, 9*6=54 for the second symmetry, leaving 444 options. Divide by 12 and we get 37 options.
  
  Overall sum: 3+3+7+33+9+37=92.
  Annotated quote matching the groups to the previous solution:
  

  rrrrrr, gggggg, bbbbb - 3, unicolor
  rrrrrg - 1 - 6 - mirror symmetry through beads
  rrrrgg rrrgrg rrgrrg - 3 - 18 - mirror symmetry through edge, mirror symmetry through beads, 180 degree symmetry (6 each)
  rrrggg rrgrgg rgrgrg - 3 - 9 - mirror symmetry through beads, no symmetry, 120 degree symmetry (3 each)
  For 2 different kinds of beads, 33, for at most 2, 36.
  
  rrrrgb rrrgrb rrgrrb - 3 - 9 - no symmetry, no symmetry, mirror symmetry through beads (3 each)
  rrrggb rrgrgb rgrrgb grrrgb rrggrb rgrgrb - 6 - 36 - no symmetry, no symmetry, no symmetry, mirror symmetry through beads, no symmetry, mirror symmetry through beads (6 each)
  rrggbb - 1 - 1 - no symmetry
  rgrgbb rggrbb - 2 - 6 - no symmetry, mirror symmetry through edge
  rgrbgb - 1 - 3 - mirror symmetry through beads
  rgbrgb - 1 - 1 - 180 degree symmetry

 

[Demystifier]

1. Extend the following to a valid equation, using only mathematical symbols!

Example: $1\; 2\; 3 \;=\; 1 \longrightarrow - (1 \cdot 2) + 3 = 1$. Solutions are of course not unique.

$9\;9\;9\;=\;6$
$8\;8\;8\;=\;6$
$7\;7\;7\;=\;6$
$6\;6\;6\;=\;6$
$5\;5\;5\;=\;6$
$4\;4\;4\;=\;6$
$3\;3\;3\;=\;6$
$2\;2\;2\;=\;6$
$1\;1\;1\;=\;6$
$0\;0\;0\;=\;6$

The easiest way to solve all such problems is to use the successor function $S$ of Peano arithmetic. For instance, the first one can be solved as
$$9-9+9=S(S(S(6)))$$
and the last one
$$S(S(0))+S(S(0))+S(S(0))=6$$

 

[mfb]

If you accept mathematical functions like that, you can also use the Heaviside step function and make it even simpler. $H(9)H(9)H(9)=H(6)$. Works for all combinations of positive integers.

 

[fresh_42]

23. John wants to buy sweets for his party. He has $100 to spend. A box Belgian pralines costs $7, chips cost $3, and for 50 cents he gets a chocolate bar. What does he have to buy, if he wants at least one of each and spend the whole $100 to buy exactly 100 pieces in total?

 

[mfb]

Is this missing something? There are many options that are trivial to find.

 

[lpetrich]

Spoiler

• Box of pralines: $7
• Bag of chips: $3
• Chocolate bar: $0.50

They must all add up to $100, and one should buy at least one of each.

I will estimate the minimum number of items purchased using a "greedy" algorithm. First, the minimum purchases, one of each: $7 + $3 + $0.50 = $10.50, giving $89.50 left over. One can have at most 12 boxes of pralines, costing $84 and giving $5.50 left over. One can have at most one bag of chips, giving $2.50 left over. One can have 5 chocolate bars, costing $2.50, with $0.00 left over.

Thus, one should purchase 13 boxes of pralines, 2 bags of chips, and 6 chocolate bars.

Relaxing that minimum-number constraint means that the solution is no longer unique, since

• 1 box of pralines = 2 bags of chips + 2 chocolate bars
• 3 boxes of pralines = 7 bags of chips
• 1 box of pralines = 14 chocolate bars
• 1 bag of chips = 6 chocolate bars

 

[fresh_42]

Is this missing something? There are many options that are trivial to find.

Sorry, yes, forgotten a condition. Corrected now.

 

[fresh_42]

Spoiler

• Box of pralines: $7
• Bag of chips: $3
• Chocolate bar: $0.50

They must all add up to $100, and one should buy at least one of each.

I will estimate the minimum number of items purchased using a "greedy" algorithm. First, the minimum purchases, one of each: $7 + $3 + $0.50 = $10.50, giving $89.50 left over. One can have at most 12 boxes of pralines, costing $84 and giving $5.50 left over. One can have at most one bag of chips, giving $2.50 left over. One can have 5 chocolate bars, costing $2.50, with $0.00 left over.

Thus, one should purchase 13 boxes of pralines, 2 bags of chips, and 6 chocolate bars.

Relaxing that minimum-number constraint means that the solution is no longer unique, since

• 1 box of pralines = 2 bags of chips + 2 chocolate bars
• 3 boxes of pralines = 7 bags of chips
• 1 box of pralines = 14 chocolate bars
• 1 bag of chips = 6 chocolate bars

Sorry, I had forgotten the condition of 100 pieces in total.

 

[kith]

23. John wants to buy sweets for his party. He has $100 to spend. A box Belgian pralines costs $7, chips cost $3, and for 50 cents he gets a chocolate bar. What does he have to buy, if he wants at least one of each and spend the whole $100 to buy exactly 100 pieces in total?

Spoiler

Let n_p, n_c and n_b be the number of boxes of Belgian pralines, of chips and of chocolate bars.

The two conditions can be expressed as a system of linear equations
<br /> 7 n_p + 3 n_c + 0.5 n_b = 100\\<br /> n_p + n_c + n_b = 100<br />
Substracting the second line from two times the first leads to
<br /> 13n_p + 5n_c = 100<br />
Since n_c needs to be a natural number, 13⋅n_p needs to be divisible by 5. This is only possible if n_p is equal to 5 (or 0 but this is ruled out by having at least one of the different sweets*). Putting it into the equation yields n_c and using the second equation from above yields n_b:
<br /> n_p = 5, n_c = 7, n_b = 100-5-7 = 88\\<br />
*Initially, I wrongly included n_p=0 as a valid solution

 

[fresh_42]

What we get are the two solutions
<br /> n_p = 5, n_c = 7, n_b = 100-5-7 = 88\\<br /> n_p = 0, n_c = 20, n_b = 100-20 = 80<br />

The second one isn't a solution as it violates a requirement.

 

[kith]

The second one isn't a solution as it violates a requirement.

Right, I thought it violated the spirit of the excercise but didn't read carefully enough! I've edited my solution.

 

[fresh_42]

Time for some fun. The following puzzle is along the lines:
Q: 1000 G in K
A: 1000 grams in a kilogram.
It has certainly no unique solutions, I know. Anyway, let's see what you have:
(Let's restrict everyone to 2 answers within the first three days, tuesday included! GMT)

24.

1. 26 L in A
2. 7 W of W
3. 12 Z in Y
4. 8 P in S
5. 27 A to C
6. 0 C of W
7. 18 H on G
8. 90 D in R
9. 4 Q in Y
10. 24 H a D
11. 2 W on B
12. 11 P in F
13. 29 F in L
14. 52 C in P
15. 64 S on C
16. 5 F on H
17. 28 S in E

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

 

[lpetrich]

23. John wants to buy sweets for his party. He has $100 to spend. A box Belgian pralines costs $7, chips cost $3, and for 50 cents he gets a chocolate bar. What does he have to buy, if he wants at least one of each and spend the whole $100 to buy exactly 100 pieces in total?

I will solve it with the correction.

Spoiler

Call the number of boxes of pralines $n_p$, the number of bags of chips $n_c$, and the number of chocolate bars $n_b$. The two conditions are
$$ \eqalign{ 7n_p + 3n_c + \frac12 n_b & = 100 \\ n_p + n_c + n_b & = 100} $$
Multiply the first equation by 2 and subtract the second equation. That gives
$$ 13 n_p + 5 n_c = 100 $$
Since all the n's are integers, $n_p$ must be divisible by 5, and its only possible positive value that also let's $n_c$ be positive is 5. This makes $n_c$ be 7, and these two values in turn make $n_b$ be 88.

Thus, John must buy 5 boxes of Belgian pralines, 7 bags of chips, and 88 chocolate bars.

 

[fresh_42]

5 3 6 4 9 2 7 5 5 3 2 8 8 4
6 8 5 7 2 9 4 6 6 8 9 3 3 7
3 7 6 4 9 2 3 7 5 6 2 8 4 9
8 4 5 7 2 9 8 4 6 5 9 3 7 2
5 2 3 9 6 3 6 8 2 4 7 3 8 6
6 9 8 2 5 8 5 3 9 7 4 8 3 5
4 6 2 3 4 5 8 4 5 7 2 8 2 4
7 5 9 8 7 6 3 7 6 4 9 3 9 7

25. Which is the fastest way to add all numbers, and what is the result?

 

[Orodruin]

Spoiler

Noting that you have grouped them (vertically) in pairs adding up to 11 and counting the number of such pairs.
4*11*14 = 616

Although ... Technically the fastest way is to copy-paste the table into Matlab and have it sum the numbers for you as this is what I started with and nobody has answered yet ... This saves all the time spent on noticing the pattern.

 

[fresh_42]

Technically the fastest way is to copy-paste the table into Matlab and have it sum the numbers for you as this is what I started with and nobody has answered yet ... This saves all the time spent on noticing the pattern.

But Matlab won't get you the Nobel prize for finding the GUT - pattern recognition might!

 

[fresh_42]

26. In a tin are 10 dimes and 5 quarters. Against a stake one can shake out 3 coins, which one can keep.

What stake is the offer worth?

 

[mfb]

But Matlab won't get you the Nobel prize for finding the GUT - pattern recognition might!

Theorists use Matlab a lot.

In a tin are 10 dimes and 5 quarters. Against a stake one can shake out 3 coins, which one can keep.

As not everyone will be familiar with the names: A dime is 10 cent, a quarter is 25 cent.

 

[DavidSnider]

26. In a tin are 10 dimes and 5 quarters. Against a stake one can shake out 3 coins, which one can keep.

What stake is the offer worth?

Spoiler: spoiler

(((2/3) * 10) + ((1/3) * 25)) * 3 = 45

 

[fresh_42]

27. An experimental setup consists of a square plate divided into 10x10 smaller squares. Exactly nine of these 100 squares are occupied by a mold. The fungus may spread to a new square Q if at least two of the four orthogonal neighbors of Q are already infected.

Can the entire 10x10 plate be occupied by mold? Why? Why not?

Hint: Consider the perimeter.

 

[fresh_42]

28. Nine cards with numbers one to nine are open in the middle of the table. Two players draw alternatively a card. Winner is who first has a hand that totals fifteen. Is there a winning strategy for the first player, for the second, or will optimal strategies always end with a draw? Why?

 

[lpetrich]

Spoiler: #28

Optimal play for both sides will always end in a draw.

With the complete set of cards, a win in two is possible, so #1 chooses one of the cards that makes such a win possible: any of 6, 7, 8 and 9. This is because 6+9 = 7+8 = 15. I select 9 for definiteness.

If #2 does not play 6, then #1 will play that card on the next turn and win. So #2 plays 6.

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.

If #2 does not play 7, then #1 will play that card on the next turn and win. So #2 plays 7.

This leaves cards 1, 2, 3, 4, and 5. Their sum is 15, but each player will play only 2 or 3 of the cards. Since none of the pairs or triplets of them will add up to 15, the game will be a draw.

 

[Orodruin]

28. Nine cards with numbers one to nine are open in the middle of the table. Two players draw alternatively a card. Winner is who first has a hand that totals fifteen. Is there a winning strategy for the first player, for the second, or will optimal strategies always end with a draw? Why?

The rules are not clear to me. Should the sum of all your cards be 15 or the sum of a subset of your cards? If the former, then you have at best drawn when your opponent forces you above 15. It is easy as player one to force your opponent above 15 and as player two to block all possible wins (as there will always be at most one single card that brings player one to exactly 15). If the latter, there is an obvious winning strategy (P1:6, P2:9, P1: 8, P2: 1/7, P1: 7/1 and win 8+7 or 6+8+1)

I also think the word you are looking for is ”alternatingly”?

 

[fresh_42]

The rules are not clear to me. Should the sum of all your cards be 15 or the sum of a subset of your cards? If the former, then you have at best drawn when your opponent forces you above 15. It is easy as player one to force your opponent above 15 and as player two to block all possible wins (as there will always be at most one single card that brings player one to exactly 15). If the latter, there is an obvious winning strategy (P1:6, P2:9, P1: 8, P2: 1/7, P1: 7/1 and win 8+7 or 6+8+1)

I also think the word you are looking for is ”alternatingly”?

The dictionary plus Google translate both said 'alternative'.

The total sum of one hand has to be exactly 15, no subsets. @lpetrich's solution is correct. The easiest way to see that a draw can be forced is:

Imagine the cards are arranged as a magic square:

4 9 2
3 5 7
8 1 6

Now it is tic-tac-toe.

 

[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.