Breaking chocolate bar into small squares

A simple solution is this. At the beginning we have only 1 piece and we have to end up with 64 pieces. Since we make only one extra piece with each cut, we have to cut the bar $\boxed{63}$ times.

You must snap the original grid seven times, and then snap each of the 8 rows/columns 7 times. This gives us $7+8\times 7 = \boxed{63}$

If you notice the pattern, you might conjecture

If you have an $n\times n$ chocolate grid, then you will hear $n^2-1$ snaps to separate the bar into $1\times 1$ squares.

Indeed, we can prove this. Using the same logic before, but only generalized: $f(n)=(n-1)+n\times (n-1) = n^2-n+n-1=\boxed{n^2-1}$

Trying to avoid complications, break the 8 by 8 bar into half, ie two 4 by 8 pieces. That's one snap.

Then, break the two 4 by 8 pieces each into half, that's four 4 by 4 pieces. Two more snaps occurred here.

2 + 1 gives 3 so far.

For each 4 by 4 piece, it's 15 snaps.

15 X 4 gives 60.

60+ 3 gives 63, and there you have it.

2^n - 1 snaps where n is the n*n dimension of the chocolate bar

Another easier solution is that a 8×8 bar will be four 4×4 bars, after 3 snaps, and each of those 4x4 bars will be 15 snaps. So total 3 + (15×4) =3 +60 = 63

3, 8, 15... We can notice that we added 5 then 7 then it must be 9 etc. 3, 8, 15, 24, 35, 48, 63, 80... the 8th term is 63.

The general solution of the number of snaps for breaking $n\times n$ chocolate bar to $1\times 1$ pieces is $n_{snaps} = n^2-1$ . For $n=8$ , $n_{snaps} = 8^2-1 = \boxed{63}$ .

Snap the 8x8 in half and each half in half = 3 snaps

Now you have four 4x4 pieces

One 4x4 takes 15 snaps

Therefore 4 x 15 + 3 = 63

If there are n rows given then first we will break all rows by making n-1 snaps and then each column can be split by making n-1 snaps, thus for n rows we will make n (n-1) snaps. Therefore total snaps = n (n-1) + (n-1) = n^2 -1. In this case n =8 So number of snaps = 8^2 -1 = 63

You can break the 8x8 piece into four 4x4 pieces using 3 snaps. The value to break one 4x4 piece is given in the question (15 snaps). Hence, you need $4*15+3 = 63$ snaps

if we split 8 × 8 into three chocolate bar as 7 × 7 and 1 × 8 and 1 × 7 individually,we hear 2 snaps then, recursion process to resolve problem which split 7 × 7 we hear the 7+8 = 15 times snap when we split into three chocolate bar as 7 × 7 and 1 × 8 and 1 × 7, include split 1 × 8 and 1 × 7 to 1 × 1 into chocolate bar 1 × 1 pieces.

def split(n): if n == 2: return 3 return split(n-1)+2*n-1 print(split(8)) 63

Each round of 3 snaps is capable of dividing a square chunk into even quarters. 3 for the first round plus 4 times 3 for the remaining 4 4x4 blocks, then iteratively again for each of the 16 remaining chunks gices 63. 3+(4 3)+((4^2) 3)=63

A simple Solution is to subtract 1 from the total of the the chocolate bar e.g 8*8=64-1=63 and it also works with other different shaped bars.

A strange solution I found is that if you add one to one side of the multiplywler and subtract one from the other side, it will always work. For instance if you have a 5x5, you adjust it to 4x6=24. I don't know why this works. Maybe soneone can explain?

Given the conditions, for NxN bar, the number of snaps could be shown through the formula (N-1) + N(N-1).

Note that for the NxN bars, the number of columns is N. With this, we can have (N-1) snaps done horizontally and it will give us N pieces of bars wherein each has 1xN dimension. To split all into 1x1 pieces, each bar of 1XN dimension will be cut into 1x1 pieces with N-1 snaps done vertically. Note that we have N bars of 1xN dimension. So, the number of snaps done vertically is Nx(N-1).

Note that the total number of snaps in NXN bar is the sum of snaps done horizontally and vertically.

Horizontally---- N-1 snaps

Vertically------- Nx(N-1) snaps

Thus, the total number of snaps in NxN bar is

(N-1)+N(N-1).

So, for 8x8 chocolate bar, we have

(N-1)+N(N-1)

=7+8(8-1)

=7+8(7)

=7+56

=63 snaps

Y n = (X o-X n) x (X o+X n) + Y o

For X o = 2, Y o = 3, and X_n = givenDimension

I got the answer through (n-1) * (n+1).

When there are 64 squares, the answer is 7*9.

2x2 (=4) -> 3 snaps

3x3 (=9) -> 8 snaps

4x4 (=16) -> 15 snaps

... finding the pattern

n^2 - > n^2 - 1

... so, it can only be 63

8x8 (=64) -> 63 snaps

(L×w)-1= # of snaps Ex (3×4)-1=11 snaps (4×5)-1= 19 snaps

Test it

Breaking a 2×2 chocolate gives us 3 snaps (2×2)-1 Breaking a 3×3 chocolate gives us 8 snaps (3×3)-1 Breaking a 4×4 chocolate gives us 15 snaps (4×4)-1 Therefore, breaking a 8×8 chocolate will give us (8×8)-1 = 63 snaps

One way to break up a chocolate bar is to snap it in half, and snap its halves in half and so on.

This is binary recursive function with $log_{2}(8\cdot8) = 6$ steps where each step contains twice as many snaps as the previous one $1, 2, 4,...$

Counting the number of snaps, you'll get the geometric progression $1 + 2^1 + 2^2 + ... + 2^{6-1}$

Which will give you $S_{6}= 1 \cdot \frac{1-2^6}{1-2} = 63$

We have 4 grids of 4x4 in a 64 square grid, we already know that for a 4x4 grid it takes 15 snaps so 4x15 for all 4 grids. Now we only need to add the number of snaps to divide the 64 grid into 4 separate grids which is 3. Therefore: 4x15+3=63 snaps for the total

Each piece make one snap, except the final one

2x2 = 4➡️ 4-1 (1 is the final piece) = 3 snaps

3x3 = 9 ➡️ 9-1 = 8 snaps

4x4 = 16 ➡️ 16-1 = 15 snaps

...

8x8 = 64 ➡️ 64 pieces - the final piece = 63 snaps

when there is a 2x2 bar then there are 3 snaps ( 2^2)-1 when there is a 3x3 bar then there are 8 snaps ( 3^3)-1 when there is a 4x4 bar then there are 15 snaps **( 4^4)-1

thus,when there is a 8x8 bar then there are 63 snaps *( 8^8)-1 * thus the answers is 63

1. Divide the one 8x8 piece into four 4x4 pieces.. This will take 3 snaps, 1 snap for halfing it into two 8x4 pieces and then 2 snaps for further splitting into four 4x4 pieces.. Now u r given that 15 snaps are required for reducing a 4x4 piece to 1x1. Since u have 4 pieces so 60 snaps plus 3 initial snaps gives u a total of 63 snaps.

If we divide it into four pieces then we get 4×4 type squares Now it is given that a 4×4 square needs 15 sounds So the answer is 15×4+3 Which is 63

$n^{2} - 1$ is the formula ( where $n$ is one side) to calculate the number of snaps!

So, here $n = 8$ .

$\therefore$ , number of snaps $=$ $8^{2} - 1$ .

Which is $64 - 1$ $=$ $\boxed{63}$

we can find thats n ^2 -1

In general...make square of 2,3,4...and then subtract 1....

Let $f(m)$ donates the snaps you will hear when a $m \times m$ chocolate bar is split into $1 \times 1$ pieces, then

$f(2^n) = 4 \times f(2^{n-1}) +3$

by the observation that we can split a $2^n \times 2^n$ chocolate into 4 pieces of $2^{n-1}\times2^{n-1}$ chocolates by 3 snaps. thus

$f(8) = 4 \times f(4) + 3 = 4 \times 15 + 3 = 63$

It takes $n-1$ snaps to break an $n \times n$ square into $n$ $1 \times n$ rectangles. Furthermore, $n-1$ snaps break each rectangle into $n$ $1 \times 1$ squares. Hence, the number of snaps to break an $n \times n$ square into $1 \times 1$ squares is $n-1 + n(n-1) = n^2 -1$ . In particular, if $n = 8$ , the number of snaps is $63$ .

You need to snap it three times to break it into 4 pieces. Then you need to snap each of those three times to break each block into 4 new pieces. Now you have 16 piece, each of which need 3 more breaks. So, 3 + 3 4 + 3 16.

64-1=63, cmon now

Simple the bar is 8x8 pieces so 8x8-1

the second assumption - that you can break either horizontaliy or verticaly - make the problem imposible ! look at the 2x2 example - after the first cut - say vertical - you have to cut horizontally and vice versa !

For every increase in n^2 you add the sequential prime number to the total number of snaps from previous answer. 2x2= 3 snaps...3x3= 3+5= 8 snaps.... 4x4= 8+7=15 snaps

Solve the recurrence equation $T (n) = 4T(\frac {n}{4})+3$