New answers tagged algebra
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 ...
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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