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

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KnotTheorist

Well, four times 17, not three, but yes, this is the only solution.
Yes, sorry. In that case it will be 15 combinations.

fresh_42

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114. How many matches need to be removed at least, in order to destroy all squares?

D115

mfb

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I can prove that it must be at least 9, but I only find (many) solutions with 10, two of them are below.
Proof: You need to destroy 16 small squares, taking away one piece can destroy at most 2. But if you remove 8 matches that each destroy two small squares then the largest square from the perimeter is still there.

fresh_42

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It can be done in $9$.

mfb

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Found a solution.

fresh_42

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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

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mfb

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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

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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

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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

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

D117

fresh_42

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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

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fresh_42

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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

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118:
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

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2018 Award
118:
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

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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

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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

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mfb

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119:
Is there a winning strategy for one of the players?
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?

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jbriggs444

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

fresh_42

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

fresh_42

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Is there a somewhat easy strategy to follow?
Yes.
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.

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mfb

fresh_42

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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

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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

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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

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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

Homework Helper
Fresh said:
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 ?)

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