Riddles and Puzzles: Extend the following to a valid equation

[fresh_42]

119. A deck of 3,040 cards is shuffled and both players, Alice and Bob, get $1,520$ cards. Each card has a number printed on it, from $1$ to $3,040$. The first player who can lay down a card such that the sum of all played cards is divisible by $3,041$ wins the game. Alice begins.

Is there a winning strategy for one of the players?

D119

 

[mfb]

119:

Is there a winning strategy for one of the players?

Spoiler

The sum of all cards is 3040*3041/2 = 1520*3041 and divisible by 3041. At the very latest the second player wins with their last card. We have a game with perfect information and no randomness (after dealing the cards) that has a finite length cannot end in a draw. For every way the cards are shuffled one of the players must have a winning strategy.

Yeah, while technically an answer it feels like cheating.

We can group the cards in 1520 pairs that each sum to 3041.. If A and B have one card of each pair then B wins with his first card. In general A must start with a card where she has the partner.
As only the sum mod 3041 matters we can relabel 1521...3040 to -1520...-1. A starts with card a1 where she also has -a1. B responds with b1 where -b1-a1 must not be in A's hand (mod 3041 of course). A responds with a2 such that -b1-a1-a2 is not in B's hand. Hmm... not seeing a long-term strategy yet.

Is there a somewhat easy strategy to follow?

 

[jbriggs444]

#119
With only 3004 cards dealt out of 3040, this is not a perfect information game.

 

[fresh_42]

#119
With only 3004 cards dealt out of 3040, this is not a perfect information game.

Sorry, typo.

 

[fresh_42]

Is there a somewhat easy strategy to follow?

Yes.

Spoiler

Alice cannot win with her first move. After that, there will always be exactly one card to win the game, and Bob can see in his cards, whether Alice has it or not. Since he has one more card than Alice, Alice can't have the winning card for all of Bob's cards, hence there is always a card he can play accordingly. Since, as you mentioned, the last card will definitely win, Bob has a winning strategy.

 

[fresh_42]

120. We have $64$ hectares of a square forest, say they are numerated like the chess board. At twelve o'clock a fire starts on $D1$. Firefighters can secure one hectare from burning every hour. Every next hour, the fire spreads to all horizontally or vertically neighboured hectares (no diagonals). How many hectares can be saved at best?

D120

 

[fresh_42]

121. Let $x <y$ be two positive integers. We have two transformations $S(x,y)=(x+1,2y)$ and $T(x,y)=(2x,y+1)$. If we start with such an arbitrary pair $(x,y)$, is there always a number $z$ and a path such that $$S^{n_1}T^{m_1}S^{n_2}T^{m_2}\ldots S^{n_k}T^{m_k}(x,y)=(z,z)$$
for some $n_i,m_j \in \mathbb{N}_0$?

D121

 

[mfb]

120. We have $64$ hectares of a square forest, say they are numerated like the chess board. At twelve o'clock a fire starts on $D1$. Firefighters can secure one hectare from burning every hour. Every next hour, the fire spreads to all horizontally or vertically neighboured hectares (no diagonals). How many hectares can be saved at best?

D120

If I understand the question correctly then the best I found is covering the E column one by one, saving half of the field. If you let it go into E then it will spread both vertically and towards the larger columns and you can only keep one side from spreading in both directions.

 

[fresh_42]

If I understand the question correctly then the best I found is covering the E column one by one, saving half of the field. If you let it go into E then it will spread both vertically and towards the larger columns and you can only keep one side from spreading in both directions.

You can save more than 50% for burning.

 

[BvU]

it will spread both vertically and towards the larger columns

Am I to understand it does NOT spread to the smaller columns ? Please clarify this spreading (e.g. after D1 it can spread to D2 and to E1 -- not to C1 ?)

 

[fresh_42]

Am I to understand it does NOT spread to the smaller columns ? Please clarify this spreading (e.g. after D1 it can spread to D2 and to E1 -- not to C1 ?)

I don't know what you are quoting here, I said

the fire spreads to all horizontally or vertically neighboured hectares (no diagonals).

So $D1\to C1, D2, E1$ or $F3 \to F2,F4,G3,E3$.

 

[mfb]

Improved fire fighting plan:

Spoiler

Make two "stairs". Save D2, E2, C3, F3, B4, G4, A5, H5 in that order. The fire needs two cycles to danger the next step in the stair, that means you can fight on both sides at the same time. A total of 20 fields will burn down.

 

[fresh_42]

Improved fire fighting plan:

Spoiler

Make two "stairs". Save D2, E2, C3, F3, B4, G4, A5, H5 in that order. The fire needs two cycles to danger the next step in the stair, that means you can fight on both sides at the same time. A total of 20 fields will burn down.

 

[fresh_42]

122. You want to put groceries into the drawer. Between three round glasses of peanut butter is still some room left for another small bottle of olives. If the radius of the peanut butter glasses is $R$ and the radius of the glass of olives is $r$, how big may be the ratio $r/R$ at most, such that the olives fit in?

D122

 

[KnotTheorist]

Spoiler: 122 spoiler

Constructing an equilateral triangle by connecting the centers of the peanut butter jars, we need to find the distance between the center of the triangle and one of the corners. This will be equal to R+r.
Thus using basic trig, we get that r = R(2/√3 -1). Thus r/R = (2/√3 -1)

 

[fresh_42]

123. Show that there are infinitely many $11-$tuples of consecutive natural numbers such that none of them is prime.

D122

 

[mfb]

Spoiler: 123

Consider x to x+10 where x=2310n+2 for a positive integer n (2310 = 2*3*5*7*11).
x, x+2, ..., x+10 are even, x+1 is divisible by 3, x+3 is divisible by 5, x+5 is divisible by 7, x+7 is divisible by 3, x+9 is divisible by 11. This gives an 11-tupel without primes for every positive n.

Alternative proof: If there would be only a finite set then the asymptotic prime density would have to be at least 1/11, in contradiction to that density going to zero.

Both proofs can be extended to arbitrary length of the tuples.

 

[fresh_42]

124. Two fair dice are rolled each round; the result of a throw is the product of the thrown numbers. A game goes over 5 rounds.

Alice throws 5 more points in the second round than in the first round, 6 less in the third round than in the second, 11 more in the fourth round than in the third, and 8 less in the fifth round than in the fourth.

How many points did she score in each of the 5 rounds?

D123

 

[KnotTheorist]

124.
and 8 less in the fourth round than in the fourth.

D123

One of these should say fifth, which is it?

 

[fresh_42]

One of these should say fifth, which is it?

Sorry. It's corrected now.

 

[fresh_42]

125. Find an integer which can be written as a sum of squares in three different ways.

D124

 

[lpetrich]

Doing #124

Spoiler

The first step is to collect the numbers. The problem statement is R2 = R1+5, R3 = R2-6, R4 = R3+11, R5 = R4-8.

We must now find dice-roll values that satisfy all these equations. The complete list of possible dice-roll values is (1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36).

The possible values of (R1,R2) are ((1,6), (3,8), (4,9), (5,10), (10,15), (15,20), (20,25), (25,30)).

The possible values of (R1,R2,R3) are ((3,8,2), (4,9,3), (5,10,4), (10,15,9), (25,30,24)).

The possible values of (R1,R2,R3,R4) are ((5,10,4,15), (10,15,9,20))

The possible values of (R1,R2,R3,R4,R5) are only one: (10,15,9,20,12). That is the solution.

 

[KnotTheorist]

125. Find an integer which can be written as a sum of squares in three different ways.

D124

Spoiler: 125 spoiler

325 = 10² + 15² = 18² + 1² = 6² + 17²

 

[fresh_42]

126. One of us has the least odd value under Euler's $\varphi$ function different from $1$, and one of us is the least that doesn't occur in its range. The third of us tells how many colors are sufficient to color a map on a ring. Me? This is the product of them.

D124

 

[fresh_42]

127. Someone replaced the round shooting targets at the local gun club by equilateral triangles with a side length of 10 inches. Show that among five hits will always be two closer to each other than five inches.

D125

 

[BvU]

129 Make a drawing !

Spoiler

Worst case for 1, 2, 3 is in the corners. dashed radii are < 5" (by at least half a bullethole size) So equilateral triangle ABC has sides < 5" and you can't find two points in there that are > 5"apart

 

[fresh_42]

128. The riches of the country are redistributed after the revolution. Nine subordinates and the former king - a total of ten people - can distribute 10 thalers per month. So every person gets a gold thaler paid out every month - even the king.

The distribution key may be changed if there is a majority for it. A subordinate agrees to a change, provided this increases his own monthly salary. He votes against it when his wages fall. If his monthly salary does not change, he abstains. The king is not allowed to vote, but he is the only one who can suggest how to change the distribution.

What is the maximum monthly paycheck the king can secure$^*)$?

Additional question: The people consists of a total of 1000 people including the king. Each month, 1000 thalers are paid, at the beginning exactly one to each person. Which maximum sum is achievable$^*)$ for the king now?

$^*)$ This means in the long run.

D125

 

[jbriggs444]

Spoiler: 128

7 and 997

Every successful change of distribution which begins with at least two subordinates having a non-zero distribution must also end with at least two subordinates having a non-zero distribution. If the result were to have only one subordinate with a non-zero distribution then there could have been at most one vote "for" and there would also have been at least one vote "against".

Consider a final redistribution that gives the king his maximum share. There must have been at least two subordinates receiving non-zero distributions. If both are victimized then three subordinates must vote in favor of the measure -- three thalers which the King cannot receive. If only one is victimized then at least two other subordinates must vote in favor of the measure. Again there are at least three thalers which the King cannot receive.

So we have an upper bound of n-3 = 7 thalers to the King.

What remains is to determine whether this upper bound is attainable. The following scheme will serve.

The first vote hands the King's thaler to subordinate number 1 and subordinate n's thaler to subordinate number 2.

After this, it is straightforward to take any subordinate with two or more thalers and redistribute them to two other subordinates, thus finally concentrating the wealth on at most two subordinates. Let n be the total distribution. Then these subordinates have k and n-k thalers respectively. Assume n is large enough that both n and n-k are greater than or equal to two (easily achievable for n=10 or n=1000).

The final moves go like:

K  S1 S2 S3 S4
0   k n-k 0 0
k-2 0 n-k 1 1
n-4 0 0   2 2
n-4 1 0   0 3
n-3 2 1   0 0

 

[mfb]

126. One of us has an odd value under Euler's $\varphi$ function, and one of us doesn't occur in its range. The third of us tells how many colors are sufficient to color a map on a ring. Me? This is the product of them.

D124

I'm confused by that one. $\varphi(x)$ is odd for x=1 and x=2 only, giving us two choices for the first number. No odd value apart from 1 occurs in its range (and various even numbers don't occur either), giving an infinite set of choices. The torus needs 7 colors.
That's a lot of options for the product.

 

[fresh_42]

I'm confused by that one. $\varphi(x)$ is odd for x=1 and x=2 only, giving us two choices for the first number. No odd value apart from 1 occurs in its range (and various even numbers don't occur either), giving an infinite set of choices. The torus needs 7 colors.
That's a lot of options for the product.

Yeah, I should have added smallest number which doesn't occur in the range, and excluded $1$. As $1$ wouldn't contribute to the product, I thought I could leave it.

 

[fresh_42]

129. A bag contains $111$ marbles in the colors red, green, blue, and yellow. If we take out $100$ marbles, we will have at least one marble of each color in our hand.

How many marbles do we have to take out at least, such that we will safely hold at least one marble of three colors in our hands?

D126

 

[BvU]

129

Spoiler

Each colour has at least 12 marbles in the bag. We can miss one colour and 11 from a second colour and still get three colours if we grab 88 marbles...

 

[fresh_42]

130. All, but less than 100 party guests, are distributed into teams of equal prime size for a game. Someone observed that the number of teams, except his, is also divisible by that prime. How many persons attended the party, if we assume that the observation is independent of the prime chosen?

D126

 

[mfb]

Spoiler: 130

Let the team size be p. There are k groups. N=kp<100. We know that k-1 is divisible by p, let's write this as k-1=cp. Then N=(cp+1)p. As 100>N>p² the only possible primes are 2, 3, 5 and 7.

For p=7 the only option is c=1, we have 8 teams of 7 members each, 56 members.
For p=5 we have the options c=1,2,3, leading to 6,11,16 teams of 5 members each, 30, 55, 80 members.
For p=3 we get 4, 7, 10, ..., 31 teams of 3 members, 12, 21, 30, ..., 93 members.
For p=2 we get 3,5,...,49 pairs, 6,10,...,30,...,98 members.

There is no prime power, therefore all party sizes have a choice how to split them into prime groups. While 66 appears in both p=2 and p=3 you could split 66 people into 6 groups of 11 members.
To split 30 in teams with a prime number of members we can only make teams of 2, 3 and 5, in all cases someone can make the described observation. That must be the solution.

 

[fresh_42]

To split 30 in teams with a prime number of members we can only make teams of 2, 3 and 5, in all cases someone can make the described observation. That must be the solution.

Yes, and the next Giuga number is 858.