# Tag Info

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Answer: (2 + (b-c)/c) + (2 + (b-c)/c) + (1 + (a-b)/c) + (b+c)/a (of course, using integer arithmetic only. no fractions) Without loss of generality, we can assume (as George has done) a >= b >= c. The layout will be taken as (a+b) x (a+c) x (b+c) in x,y and z directions. A simple greedy stacking will try to keep axb on floor (x,y plane) as far as ...

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Answer Without loss of generality assume the board has 7 rows and 8 columns. Clearly we want all dominoes to be horizontal. Let us say that a state P is optimal if it is not possible to reduce the number of vertical dominoes with legal moves. Clearly the desired end state has zero vertical dominoes and is hence optimal. Suppose P is optimal but has at least ...

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You need to cover the four corners, so one of the smaller squares covers two corners of the bigger one and the other two smalls cover one corner each.

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With the following arrangement you can easily stack pieces into the box: I have assumed without loss of generality that $a<b<c$, but as Damien_The_Unbeliever noted in the comments, it also assumes that $a+b>c$. That does not matter however, as this arrangement can be tweaked to insert one more piece: Now it just remains to be proved that it is ...

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Almost complete answer using Jaap's lower bound, i.e. ignore my lower bound and skip the first two blocks: Now that @Jaap Scherphuis has bumped the lower bound to It remains to be shown that Amy can choose in such a way that more becomes impossible.

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I think it's Reasoning: So, we have: Which also leads to Amy's tactic:

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Observe that every 4th power is If one of $a,b,c,d$ is $9$, then If there is a $7$, then we must have $a=9$, so the other two digits are at most $4$ (from the RHS). Here there is If there is no $7$, then we must have either $a=6$ or $a=9$; since $6^4+9^4>7000$, it must be $a=9$. The LHS is over 9000, so the other three numbers must be $6,6,?$ or $6,5,5$...

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First, let us define some things: For simplicity, for partial boards presented (with ...), let's consider that the width is equals to or larger than the height. If not, you can just rotate everything 90° to get a board that is like this. Unsolvable board (UB) - One that no matter what you rotates, it is impossible to have all the dominoes with the same ...

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I think the answer is Explanation: First place one of the $4\times4$ squares within the $5\times5$ square. The remaining area (9 units) should be as compact as possible, so let's shift the $4\times4$ square right up to one corner, leaving an L-shape remaining. (I'm not sure how to prove rigorously that this is optimal.) Now we need to place the other two ...

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Is it the following? Very broadly, other than standard sudoku solving techniques:

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This is because

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This is a standard application of the Burnside Lemma. I'll solve the more general case of a square with $n$ colours.

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I think the answer is Counting

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We can just draw a cube like so: The answer is...

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This is true for a given $n$ if and only if Now, Can we make that work? Therefore Therefore The above argument may be difficult to follow. Let's look at it more concretely. Now And in fact To play the same game

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Let us denote the ages of Person 1, Person 2, Person 3 by $x,y,z$ respectively. We'll assume that $x,y,z$ are positive throughout. The product of the 1st person's and the 2nd person's ages is $311 \frac{2}{3}$ plus the 3rd person's age. The sum of the 1st person's age and the quotient of the 3rd person's and the 2nd person's ages is $41 \frac{17}{24}$ ...

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I've made this community wiki, so please edit away! This is a very good question to put to students as long as one subsequently hammers home the point that it is ill posed and one works out the common fallacies that contribute to the expected answer. The question is actually well suited for this didactic exercise because the tacit assumptions OP's preferred &...

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Just keep a running total of what each is owed (positive) or owes (negative). It doesn't matter to whom: There are people A,D,G,P "A" needs $2.50 from D,G,P "P" needs$2.50 from A,D,G As far as who paid "A" what. "D", paid "A" \$5 (So "A" owes D \$2.50) "G", hasn't paid "A&...

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I observed the following: This is because This observation immediately excludes many numbers from consideration. It remains to be shown that the numbers that were not excluded do all end at $153$. For completeness, here is my working out of the remaining cases. Rand al'Thor already did this first in his answer. Like him, I do not see any clever way that ...

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Considering cycles The largest number such a chain can ever reach is $1486$ (every number between $2001$ and $2100$ gives at most $8+0+729+729=1466$ at the first step, and the largest possibility resulting from any number up to there is $1+27+729+729=1486$). So we have an upper bound, which means every chain must eventually end in a cycle. In the OP you ...

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(After rechecking my numbers which gave me a different answer before, it seems I've written a duplicate of hexomino's quicker answer. Since this one has 2 lines of maths instead of many, I think I'm going push post anyway.) The key to figuring this out is that This allows us to figure out the relative frequency of the two cases: From this we get that ...

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For these sorts of problems I like to use In this scenario we can apply it as follows

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The most elegant solution I could find was this one: let the matrix be \begin{equation*} \begin{pmatrix} A & B & C \\ D & E & F \\ G & H & I \end{pmatrix} \end{equation*} Let the sum of each row/column/diagonal be $S$. Then \begin{eqnarray} A+B+C + D+E+F = A+E+I + C+F+I = 2S &\to& I = \frac{B+D}{2} \\ A+D+G = G+H+I + S &\...

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First I'll prove a property of $3\times3$ magic squares. Using this property you can use a similar proof to find the central cell in this case: The rest of the magic square then follows: I originally used a less elegant more general method by finding a generic solution: Now it is just a matter of applying that to this particular problem.

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Let's say A pays £10 for court for four people including himself (So £2.50 each) P pays £10 for a court for four people including himself. (So £2.50 each) That’s equivalent to A lending £2.50 to three people. And P lending £2.50 to three people. So that leads to what is mentioned at the beginning of the question "A" needs \$2.50 from D,G,P "P&... 2 Mathy If you add a "hub" node to your graph, you can reduce the number of possible connections. If you allow 2 arrows to represent "owed" and "owes", it takes the number of possible relationships down from (n-1)n to 2n (so equal at 3 nodes and smaller thereafter). Non-Mathy (the practical use of the above) You invoke the money ... 2 The initial state of who owes whom what, before any payments are made, looks like this: Owes ║ A │ D │ G │ P Total O ═╬═══╪═══╪═══╪═══ ═════ w A║ X │250│250│250 750 e ─╫───┼───┼───┼─── ───── d D║ 0 │ X │ 0 │ 0 0 ─╫───┼───┼───┼─── ───── T G║ 0 │ 0 │ X │ 0 0 o ─╫───┼───┼───┼─── ───── P║250│250│250│ ... 7 Graphs are your friends! 1: Set Up: To make matters simple, let's just assume that at one point, A loaned the other three \$2.50 and P did the same. Let a directed edge represent \$2.50. Our starting graph represents the cash flow state after both sets of loans are made. Note that the A-P Edges cancel. 2: D pays a \$5.00. Draw two edges (green, to ...

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Hey there this is my first time so I don't know how to use the system exactly, pardon me. Let's make some diagrams. To represent A needs 2.5 from D, use the text A<---2.5---D. Then we can represent the needed transactions as: A<---2.5----D A<---2.5----G A<---2.5----P P<---2.5----A P<---2.5----D P<---2.5----G First observe that there ...

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At the beginning: The only official payment that has happened is D paying 5.00 to A, so the updated transactions are: Now then:

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Assuming by "needs" at the start you mean "is owed", then: Also: So: You could work out But it's easier to just work out In this case Or another way of looking at it

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However, he only had 30 of the cheaper melons, so he could only make 10 groups containing 3 of the cheaper. So 10 groups of (3, 2) uses 30 cheap, and 20 expensive, leaving two groups of (0,5). These 10 melons should have been sold for 5 d, but he ended up selling them for 4d, losing 1d on the transactions.

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A liberal interpretation of “coincides” in the puzzle statement ... “each vertex of the triangles coincides with exactly three triangles” ... allows for a vertex to touch a side, and not always another vertex, of another triangle as in these failed attempts ... ... that led to this pair of ...

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suppose plan area is A and no of triangles are n, A = n * 1/2 * h * b so if we want n, n = 2A /(hb), that is the minimum no of triangle

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Looks like the apparent relation between the shapes of the arrow heads and the operation they imply is a red herring. And even if they aren't there is no reliable clue as to what open versus filled heads may signify.

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The number of triangles in my best solution is but I don't know if this is optimal. Addendum: I previously had an incorrect solution, as I used similar triangles instead of congruent ones (i.e. I used some triangles of a different size but the same shape). As requested by @humn, I'll keep that incorrect solution with 12 triangles available below:

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In both the square, area will be the same. suppose big square has area, A = x*x all small square total area will be A = n * 1 *1 so, x*x = n * 1 * 1 x = √n, for all x>=n

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Because

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Occam's razor says the simplest explanation is the best.

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It's simply just So Therefore the missing number is

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X could be operator (IS NOT) ! Because 1x=2 2x=4 3x=3 4x>8 are always either true or false conditions in any programming language. 1!=2 //Always True 2!=4 //Always True 3!=3 //Always False 4!>8 //Always True

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I must admit that I looked this up, but the largest graph which works turns out to be unique, and is called Now all that remains is to label the vertices appropriately. One easy way to do this is to Of course the numbers you get are rather large. You can improve this a bit as follows: There may be a way to reduce the numbers further.

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Answering whether the $N/(N + M - 1)$ survival probability can be met:

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The following 12 numbers satisfy the conditions: $$203,385,437,713,814,1330,1479,1495,2418,3441,11951,70499$$ Represented as a graph where the numbers are its vertices, two of which are joined by an edge if they have a common divisor greater than 1, the graph can be shown not to have a complete subgraph on four vertices ($K_4$): nor does its complement: In ...

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Their ages are: Reason:

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