In this image I have counted 14 but others say 18.
Is there a method to check exactly?
[Math] How many squares are in this image? Is there a method to check
puzzle
Related Solutions
A straightforward search yields the figure of $4392$ for all but the $1 \times 1$ stones. The former fill $14$ out of $20$ squares, so there are $\binom{6}{4} = 15$ possibilities to place the latter. In total, we get $$4392 \times 15 = 65880.$$ These can all be generated, and one can in principle calculate the number of connected components in the resulting graph, where the edges correspond to movements of pieces.
Edit: There are 898 different connected components. There are 25955 configurations reachable from the initial state.
$LOVES+LIVE=THERE$. Expanding it in decimal form, I have $ (L\times10^4 + O\times10^3 + V\times10^2 + E\times10^1 + S\times10^0) + (L\times10^3 + I\times10^2 + V\times10^1 + E\times10^0) = (T\times10^4 + H\times10^3 + E\times10^2 + R\times10^1 + E\times10^0)$
$\Rightarrow L\times10^4 + (O+L)\times10^3 + (V+I)\times10^2 + (E+V)\times10^1 + (S+E)\times10^0 = (T\times10^4 + H\times10^3 + E\times10^2 + R\times10^1 + E\times10^0)$
$\Rightarrow L=T \text{ }$ which is impossbile because it is already said that two letters can never represent the same number. Hence, assume that $L \neq T$. This clearly means that there is some kind of carrying operation from the preceding additions. Also,I equated the like powers and obtained the upper that every letter could have. This brought me to the following conclusions ( by equating like powers ):-
Since, $L\neq T$, we have to have a carry operation and hence the maximum possible of $T=9$ and for this we have to use the maximum possible value of $L=8$.
Now, $O+L=H$. For this, we have to calculate the maximum possible values of $O$ and $H$. This means that $O=7$ and $H=5$ because $1$ will be carried over to the higher place and hence used for this calculation of $T$.
Now, the trick is to solve for $V+I=E$.Here, we have 5 choices for $V$ and $I$ each - $1,2,3,4,6$. Here we need to perform some trial and error to eliminate some of the choices. Here, most of cases are not suitable because they lead to redundancy in the digits. These possible combinations of $(V,I)$ are $(6,4); (4,6); (6,3); (3,6); (6,2); (2,6); (6,1); (1,6); (4,1); (1,4); (4,2); (2,4); (4,3); (3,4); (3,1); (1,3); (3,2); (2,3) $. The remaining pairs are $(1,2)$ and $(2,1)$. Out of these the pair $(2,1)$ is useless because this makes $E=3$ which in turn makes $R=5$, again a redundancy. Hence, the most viable option is the pair $(1,2)$ for which $E=3$ and hence $R=4$. Also, since $S+E=E$, we can say that $S=0$ and hence the $E$ at unit's place is also $3$. Hence, our number is 95343. Hence the maximum value of THERE is 95343. Hence there are 95343 LOVES in THERE.
Best Answer
[I will give two methods. The SECOND one is the better one.]
Method? Well, sort of, you can look at each "atom" piece and count how many squares that is the upper right hand corner of.
A)Top right square -> top right square; engulf the the rectangles for a 3 by 3, the whole 4x4 = 3.
B) top middle rectangle -> 2x2 square; 3x3 square = 2.
C) top left square -> top left square = 1.
D) middle side rectangle -> 2x2 square; 3x3 = 2.
E) next square in the (2,2) spot -> 1x1;2x2;3x3 = 3
F) (3,2) spot ->1x1; 2x2 = 2
G) (4,2) spot -> 1x1 = 1
H) (3,2) spot -> 1x1; 2x2 = 2
I) (3,3) spot -> 1x1 = 1
J) (4,1) spot -> 1x1 = 1
K) low middle rectangle = 0
L) (4,4) spot -> 1x1 = 1
So total: 18
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This i a simplification of simply figuring out the squares in a complete grid and subtracting the one that rectangles make impossible.
A 4x4 grid will have: 16 1x1 squares; 9 2x2 squares (as there are 3 squares in each of the top 3 rows that can be an upper right hand corner of a 3x3 square), 4 3x3 squares, and 1 4x4 square.
So an n x n grid will have $\sum k^2$ total squares.
In this case 16 + 9 + 4 + 1 = 30.
The first top rectangle eliminates 2 1x1 squares and 1 2x2 square. So only 27 possible.
The left side rectangle eliminates 2 1x1 squares and 1 2x2 square. So only 24 possible.
The right side triangle eliminates 2 1x1 squares (1 2x2 that was already eliminated) and a 3x3. So only 21 possible.
The bottom rectangle eliminates 2 1x1 squares, 2 2x2 squares. So only 18 possible squares.