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

[fresh_42]

115. There is a cube, constituted of 9 smaller ones, like Rubik's. On all three sides of every corner cube is a same number attached, either three times $+1$ or $-1$. Finally every center field carries $+1$ or $-1$, too. The respective value is calculated as the product of the four corners that are located on the respective side surface.

Is it possible, that we choose these fourteen numbers in a way that they sum up to $0$ and how?

D116

 

[mfb]

There is a cube, constituted of 9 smaller ones, like Rubik's.

3x3x3 = 27?

Let the corners be identified as (x,y,z) where a,b,c= 0 or 1.
Let (0,0,0)=(0,1,0)=(0,0,1)=-1 and use 1 for all other corners. There is one face (x=0) showing -1 from three -1 corners, two faces (y=0 and z=0) showing +1 from two -1 corners, two faces (y=1 and z=1) showing -1 from one -1 corner and one face (x=1) showing +1 from no -1 corner. Sum: -3+5-1+2-2+1 = 0

 

[fresh_42]

3x3x3 = 27?

I avoided that since I feared endless discussions whether there is a cube at the center or not ...

Let the corners be identified as (x,y,z) where a,b,c= 0 or 1.
Let (0,0,0)=(0,1,0)=(0,0,1)=-1 and use 1 for all other corners. There is one face (x=0) showing -1 from three -1 corners, two faces (y=0 and z=0) showing +1 from two -1 corners, two faces (y=1 and z=1) showing -1 from one -1 corner and one face (x=1) showing +1 from no -1 corner. Sum: -3+5-1+2-2+1 = 0

I don't get it. There is only one number per corner, eight corners in total and six numbers at the center of the surfaces, namely the product of the four numbers in the corners, respectively.

 

[mfb]

There is only one number per corner, eight corners in total and six numbers at the center of the surfaces, namely the product of the four numbers in the corners, respectively.

That's how I understood the puzzle. I made a calculation error, however. -3+5-1+2-2+1 = 2 not 0.

Correction: It is impossible. Consider what happens if we change one corner from -1 to 1. We flip the values of all three adjacent faces. If all three where -1 before they are now all 1, we increased the sum by 8. If only two of them were -1 then we increase the sum by 4 (in total we change three -1 to 1 and one 1 to -1). If one of them was -1 then we keep the sum the same, and if they were all 1 then we reduce the sum by 4. Every corner change keeps the sum the same modulo 4. "All corners are 1" leads to a sum of 14, which has a remainder of 2 when divided by 4. All configurations are reachable via corner changes from this state, therefore they all have a remainder of 2 when divided by 4 - a sum of zero is impossible.

 

[fresh_42]

116. What is the smallest number that actually needs all nine cubes to write it as a sum of cubes?

D117

 

[fresh_42]

117. Find all $4$ even natural numbers $n$, such that $n=p+(n-p)$ is a sum of two primes for every possible prime $n/2 \leq p < n$.
E.g.: $10=5+5=7+3$

D118

 

[fresh_42]

118. If we multiply a two digit number such that the three digit result is obtained by moving a $0$ between the digits, which possibilities do we have?

D119

 

[mfb]

118:

Spoiler

We transform xy to x0y, which means we add 90*x. xy must be a factor of 90*x. Note that 90=2*3*3*5. Trivially x0 becomes x00 after multiplication by 10, I don't list these in the following:
x=1: xy=15 (becomes 105 after multiplying by 7), 18 (becomes 108 after multiplying by 6)
x=2: -
x=3: -
x=4: 45 (becomes 405 after multiplying by 9)
x=5: -
x=6: -
x=7: -
x=8: -
x=9: -

Hmm.. just casework.

 

[fresh_42]

118:

Spoiler

We transform xy to x0y, which means we add 90*x. xy must be a factor of 90*x. Note that 90=2*3*3*5. Trivially x0 becomes x00 after multiplication by 10, I don't list these in the following:
x=1: xy=15 (becomes 105 after multiplying by 7), 18 (becomes 108 after multiplying by 6)
x=2: -
x=3: -
x=4: 45 (becomes 405 after multiplying by 9)
x=5: -
x=6: -
x=7: -
x=8: -
x=9: -

Hmm.. just casework.

Yes. As several members use coding to solve problems ...

116. was very easy without the need of code. 117. has three easy solutions and one which is a bit more to check. I guess nobody has coded a primality check.

 

[mfb]

I was thinking about it, but a program to run over some small numbers feels boring.

116: 23 = 8+8+7*1
It is interesting that only small numbers require so many cubes. 23 and 239 are the only ones needing 9. 454 is the last one needing 8. We know large numbers can need at least 4 and need at most 7 but it is not known what the limit is.

 

[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.