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5

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 ...


3

Partial answer:


0

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


4

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


5

I think the answer is Counting


6

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 $41 \frac{17}{24}$ ...


4

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 ...


4

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 ...


3

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 &\...


3

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.


11

The number is because we can rewrite the three given conditions as follows, where $H$, $T$, and $U$ are the hundreds, tens, and units digits respectively: Adding the first two gives Also from the first equation, So the number must be


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