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

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#### fresh_42

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What we get are the two solutions
$$n_p = 5, n_c = 7, n_b = 100-5-7 = 88\\ n_p = 0, n_c = 20, n_b = 100-20 = 80$$
The second one isn't a solution as it violates a requirement.

#### kith

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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

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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

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#### 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.
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 lets $n_c$ 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

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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

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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

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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

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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

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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

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

#### fresh_42

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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.

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#### fresh_42

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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

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

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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

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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

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@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.

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#### fresh_42

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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

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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

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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

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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

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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

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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

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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

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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.

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.

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