I solved a puzzle about proving that if a rectangular board can be covered by L-Tetrominoes then the number of squares must be a multiple of 8. I based the solution on a colored board (like a chessboard) and 2 types of L-Tetrominoes Either:  _____ |BLACK|     |WHITE|_______   |BLACK| WHITE| --------------  or  _____ |WHITE|     |BLACK|_______   |WHITE|BLACK | --------------    The follow up puzzle is: Prove that an 8x8 board can not be covered with 1 O-Tetromino and 15 L-Tetrominoes. The O-Tetromino is:  ____ ____ |    |    |     |----+----|     |____|____|      I thought of the following proof: If I use 1 O-Tetromino then there remains a board of 8x8-4 = 60 squares. 60 is not a multiple of 8 and we have already shown that for a board to be covered with L-Tetrominoes must be a multiple of 8. Therefore the 8x8 board can not be covered with 1 O-Tetromino and 15 L-Tetrominoes Is this proof valid? It seems to informal also in relation to the approach I followed to prove that the L-Tetrominoes cover a board of x8 If it is not, then how can I prove it?
120769
Prove that an 8 x 8 board cannot be covered by 15 L-tetrominos and one square tetromino (an L-tetromino is a plane figure shown below, constructed from four unit squares arranged in the form of L; a square tetromino is a plane figure shown below, constructed from four unit squares arranged in the form of a square)
67452
The following question popped up during a conversation with a friend of mine: is there any integer $n$ such that $n$ is not a square in $\mathbb{Z}$ but it is a square in every field $\mathbb{Z}_p$ with $p&gt;n$?  It sounds like an elementary question, and I am almost sure that the answer is no (mostly because of some distant memories of mine about a theorem of Hasse and Minkowsky), but I was not able to find a definitive answer.    I tried to apply some Model Theory to it, but I don't think it can be a fruitful path to follow. Another thing I noticed is that if I can prove that there is an $a$ such that $n=a^2$ in two different fields $\mathbb{Z}_p$ and $\mathbb{Z}_q$, then the the statement must be false: otherwise, writing $$a^2=n+bp$$ and $$a^2=n+cq$$ for the smallest possible $b$ and $c$, we would have $bp=cq$, a contradiction since neither $b$ nor $c$ can be $0$. But again, I got stuck after this consideration. Do you have any ideas or suggestions?
121401
Two rooks are placed on an 8 by 8 chessboard such that they're in different squares. What is the probability they can't attack each other? Rooks can attack each other if and only if they're in the same row or column. I know that there are questions like this but they give the general numbers not the probability I am not sure how to start the questions and I have looked at this questions for hours. Any help would be welcome.
2377
Suppose we have an $2^n\times 2^n$ board. Prove you can use any rotation of L shaped trominoes and a monomino to fill the board completely.  You can mix different rotations in the same tililng.
113249
How many positive integer $n$ are there  such that $2n+1$ , $3n+1$ are both perfect squares ?   $n=40$ is a solution . Is this the only solution ? Is it possible to tell whether finitely many or infinitely many solutions exist ?
48908
Consider the squares of an $8 \times 8$ chessboard filled with the numbers $1,2,3,4 \ldots ,64$ in sequential order.   If we choose $8$ squares with the property that there is exactly $1$ from each row and exactly $1$ from each column, and add up the numbers in the chosen squares, show that the sum obtained is always $260$.  Please provide full solution and explanation.
66140
I am looking for an algorithm that can return the number of size of n squares that fit into a a rectangle of a given width and height, maximizing the use of space (thus, leaving the least amount of leftover space for squares that do not fit). Neither the rectangle nor the squares can be rotated.  For example, let's say I have a rectangle that is 5 inches by 7 inches, and I need it to fit 35 squares. The algorithm needs to tell me that the 35 squares fit if they are 1 inch wide/tall (as they could be laid out inside the rectangle in a 5 x 7 grid). Another example is if I need to divide a rectangle 35 inches by 1 inch into 35 squares. It should still tell me that the squares will fit if they are 1-inch wide/tall (as they could be laid out in the rectangle in a 35 x 1 grid).  The tricky part is that sometimes there may be leftover space, as the squares cannot be divided into partial squares. Let's say for either of the two examples above I need to lay out 34 squares and not 35 (in which case the answers would still be 1 inch), or maybe 33, or 7 squares. Or, perhaps the rectangle width and height aren't whole numbers. With the number of squares being a variable I need an algorithm that can tell me the size of the squares for a given rectangle width and height.  Thanks for your help!
4247
We have a $100\times100$ board divided into $10^4$ unit squares. These squares are coloured with four colours so that every row and every column has $25$ squares   of each colour. Prove that there are $2$ rows and $2$ columns such that their $4$ intersections are painted with different colors.   My attempt:  I first counted number of pairs $\mathcal P$ $(r_1,r_2)$ such that $r_1,r_2$ belongs to same row, but different in colour. This can be done in $$\binom 42\times25\times 25$$ ways. So, in $100$ rows, this number is $$100\times\binom 42\times25\times 25$$ Now, we can find a pair of columns in $$\binom{100}2$$ ways.  So, by Pigeonhole Principle, we can find at least a pair of columns containing $$\left\lceil\frac{100\times\binom 42\times25\times 25}{\binom{100}2}\right\rceil\\=\left\lceil\frac{100\times75\times 50}{50\times99}\right\rceil\\=76$$ of those pairs $\mathcal P$.    But unable to proceed anymore. And this approach too seems misleading, because in every pair of columns, we can always find $\binom42\times25\times25$ pairs from $\mathcal P$. So, this approach makes no sense, I think.  What should be correct approach? And also, "if we pour $nk+1$ objects into $n$ urns, then there is an urn with $k+1$ objects", can this concept be plugged in this question?
62381
Can you please help me and tell, how should I move on?  Can this be proved by induction?     Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$.   Thank you in advance
85735
I've seen two questions on solving the Rubik's cube but none of the answers have given a complete solution using mainly mathematical techniques. Furthermore, I've not seen a good explanation of general techniques for solving permutation puzzles in general, including those of the Rubik's cube family, without involving any memorized move sequences. So I've decided to write up my method below as an example of how group theory can be applied to solving such puzzles.
43854
Why can't I tile a $100 \times 100$ table with $1$ by $8$ pieces?  If we look at the number of tiles, $100^2$ is divisible by $8$. So this does not contradict existence of such tiling.  The  of coloring the squares by black and white color does not help here either, since the $1\times8$ domino has the same number of white and black squares and so does the whole board.
21780
This came up while solving another ENT problem. I want to ask when is: $$2^n -7 \text{ where } n\geq 3$$ a perfect square? Specifically, I also wanted to know what would be the solutions when $n$ is odd? How should I solve this?  I can check that $n=3, 4, 7$ are solutions but cannot find more. As it is of the form $4k+1$, it doesn't help as well.
110739
I need some help with the exercise I was assigned in math class. We were given a 3x3 square made of unit squares, and we wanted to paint all the squares. We had 3 sets of paint, red, blue, and green. Each paint was distinct from each other, but two unit squares with the same color were the same. Additionally, there's only enough paint of each color to color 3 unit squares, and all the paint has to be used. The goal of the exercise was to find the total amount of distinct possible ways such that no two unit squares that share an edge (that is to say they are next to each other vertically or horizontally, having same color across a diagonal is fine). I've started by trying to take a look at each possible arrangement of the bottom 3 unit squares, but I can't seem to find a pattern between it. Is there a better way than continuing by brute force? Any ideas on how I should proceed from here?
101233
Let $k$ be a natural number. Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.  I tried to prove this by supposing one of them is a square number and by substituting the corresponding $k$ value. But I failed to prove it. If we ignore one term, we can make the remaining terms squares.  For example, $3k+1$ and $4k+1$ are both squares if $k=56$ (they are $13^2$ and $15^2$); $3k+1$ and $6k+1$ are both squares if $k=8$ (they are $5^2$ and $7^2$); $4k+1$ and $6k+1$ are both squares if $k=20$ (they are $9^2$ and $11^2$).
120606
What is the maximum amount of checkers you can place on an $m \times n$ checkerboard such that no four checkers make a rectangle parallel to the rows and columns?   The square colors don't matter; it is essentially a grid. Also the checkers are all indistinguishable.  I tried looking at the problem row by row, but this devolved into too many cases. I've also looked at small values of $m$, but couldn't get anywhere.  It seems that if there are a lot of rows, the optimal way is to place 2 checkers for the first $\binom{n}{2}$ rows, and then place 1 continuing down the rest. However, when there are relatively few rows, there seems to be a lot of cases regarding how many checkers to place for each row.
17573
It seems to me like the answer should be $n!$. Since you can first select the first vertex, then select each remaining vertex until you've got them all in a permutation.  My textbook, however, says that the answer is $(1/2)(n-1)!$  Unfortunately, it doesn't explain why this is. Why is it?
5350
I am pretty sure this question has something to do with the Least Common Multiple.    I was thinking that the proof was that every number either is or isn't a multiple of $3, 5$, and $8\left(3 + 5\right)$. If it isn't a multiple of $3,5$, or $8$, great. You have nothing to prove. But if it is divisible by one of them, I couldn't find a general proof that showed that it wouldn't be divisible by another one. Say $15$, it is divisible by $3$ and $5$, but not $8$.