Show your working

[math]x_2 = 4\qquad (1)[/math]
[math]x_3 = 9\qquad (2)[/math]
where [math]x_n[/math] is the area of a square of side length [math]n[/math]

[math]y_2 = 5[/math]
From [math](1)[/math],
[math]y_2 = 2^2 + 1[/math]
or, more generally,
[math]y_n = n^2 + 1[/math]
[math]∴ y_3 = 3^2 + 1 = 10[/math]

[math]QED.[/math]

>he thinks there are only 10 squares
there's 14

A correction:
Instead of [math]y_2 = 2^2 + 1[/math] I meant [math]y_2 = x_2 + 1[/math].

well since 5 is the 4th Catalan number then it must be the 9th Catalan number 1430

Number of squares or proportion.

Noice

4 squares ----- 5
9 squares ----- x

4x = 5*9
4x = 45
x = 45/4
x = 11.25

Pretty sure from playing around with it that it's grids_n + squares_n-1
So grids_n = 3x3 = 9
And squares_n-1 is the numbers of squares on a 2x2 grid which is 5
To total 14.
That would make 4x4 = 16 + 14 = 30, 5x5 = 25 + 30 = 55, etc.

A nice excercise would be to put this solution into a closed form without recursion.

7

sqrt(total intersections) is the index into a set of prime numbers
9 - > 5
16 -> 7

4 1x1 squares + 1 2x2 square = 5 squares
9 1x1 squares + 4 2x2 squares + 1 3x3 square = 15

Are individual squares of equal size? Because they don't appear to be.

12

>Veeky Forums still hasn't mastered addition
kek

work:

n = number of unit segments
x = number on the right

n=(x+1)*2

So in the first one n=12 and x=5 and the second one n=24 and x=11.

11 is my answer
n =

I tried Lagrange interpolation and came up with 5. Thank me later.

oops 14 rather

still better than all these pseuds doing math until they're up their own ass instead of seeing the obvious

Its the number of even degree vertices.
So the bottom is 8.

(Number of vertical lines + Number of horizontal lines) - 1

3 + 3 - 1 = 5
4 + 4 - 1 = 7

No, wait!

I changed my mind.

It's (Number of small squares + 1)

4 + 1 = 5
9 + 1 = 10

No, wait again!

I propose it's ((Number of small squares * 3 / 2) - 1)

(4 * 3 / 2) - 1 = 5
(9 * 3 / 2) - 1 = 12.5

No, no, no!

It's actually (5 + Number of vertical lines - Number of horizontal lines)

5 + 3 - 3 = 5
5 + 4 - 4 = 5

OK, I'm getting bored.

But I wish the posts I made above show everyone how retarded OP's question is.

There are countless "solutions".

Now if you ever see a retarded question like this again it's your duty to ignore it or point out why it's retarded.

Have a nice one.

kek at the fuggin underage sperg

Based on the available information, the answer is 5.
>reasoning: there's a box on the left and a 5 on the right.

The total number of squares is the sum of the first square naturals from 1 to n, where n is the lenght (in small squares) of each individual square. For 1 is just 1 for two you have 2x2 and its 1 + 4=5 For three is 1 + 4 + 9 =14. My intuition is solid, i think, but I still lack a proof. I will report later.

Right answer. had the same idea but fucked up when adding them together.

That's most likely correct but it's still a series, I meant the closed form of the summation

en.m.wikipedia.org/wiki/Square_pyramidal_number

Neat

4 + x = 5
x = 1

8 + 1 = 9

keep trying

72
Prove that I'm wrong faggot.

my first thought was just to solve for area

A number of squares.
so it should be 14,
9 small
4 made out of 4 smaller squares
and 1 large made out of all the small squares.

14

If 4 boxes = 5 then 8 boxes should = 10

That interpretation is just as valid as any other unless there's more information given.

you always make it more difficult than it needs to be
4 squares = 5
9 squares divided by 4 = 2.5
2.5 times 5 = 12.5 - there is your answer
your welcome!

ooops should be 9 divided by 4 = 2.25
so answer is actually 11.25
your very welcome!

5 intersections between lines up top (excluding the corners)

12 in the one below

i just did this and got 11.25 for Y

wut

and who gives a shit about 8 boxes?

45/4 dumbasses.

2^2 + 1 = 5
3^2 + 1 = 10

Btw I just realized most primes squared either end in "1" or "9", is there any reason why ?

yeah, that's probably a finite field effect over mod 10

If you check, it's only the last digit in the prime that should have any affect on the last digit in the result

11,25

1->1, 2->4, 3->9, 4->6, 5->5, 6->6, 7->9, 8->4, 9->1

is there a fast square root method here?

1496569410
i forgot the zero

4= 5
9= 10

Its not. There is no reason to define the squares as x instead of, say points. The question doesnt have an answer.

>45/4 = 13.5

The answer is 10.

I though 1.25 points per rectangle, so 1.25x4 is 5
9x1.25 is 11.25
OP says show thinking, so i come out with the most basic one

2x2 cube has 12 square sides
3x3 cube has 24 square sides
It's therefore logical to conclude the answer is 5x2=10
t. engineer undergrad :^)

I don't get why so many retards say 14.

In the top pictures you have 5 faces.
In the bottom picture you have 10 faces.

In general you can use Euler's formula for that stuff.
en.wikipedia.org/wiki/Planar_graph#Euler.27s_formula

Because the picture lacks information so everyone just goes with whatever shit.

bumping for answer

2*2 cube = 5
3*3 cube is 2.25 times larger than 2*2 cube (4 squares vs 9)
5*2.25 = 11.25

Nigga, why are you writing on the floor? Can't you afford paper?

that

Number of smallest squares plus one
Question is retarded. You can make infinite patterns from a single data point

You can make infinite patterns from any number of data points

Don't you sass me Jimmy

The answer is 12.

It is a networking problem counting only nodes. Either it is drawn like shit and the corners aren't nodes or it is only nodes with some fancy name I can't fucking remember right now.

>Extrapolating from one datapoint

Let Q be "A square formed by four smaller squares: One top left (Q00), one top right (Q10), one bottom left (Q10) and one bottom right (Q11)." We know that Q = 5.

In the picture below we have 4 overlapping Q's. So if we ignored the overlap, we would have 4Q = 20.

Now, let's consider the overlap:
- The top left Q overlaps with its bottom right square (Q11).
- The top right Q overlaps with its bottom left square (Q01).
- The bottom left Q overlaps with its top right square (Q10).
- The bottom right Q overlaps with its top left square (Q00).

Now, when we combine Q11, Q01, Q10 and Q00, we get a single Q (straight from the definition).

Thus the answer is 4Q - Q = 3Q = 15.

fuck your retarded retardings

"1,2,_,_,_"
if you were ever asked to complete this pattern and it harmed you emotionally, i feel bad for you, but you can't go around the internet crying about it.

OP here. 11.25 is the correct answer. Why the fuck is Veeky Forums so bad at simple questions?

These and whoever else posted "14" as the answer are correct. If you want any other answer, you need to provide extra "rules" for the "puzzle".

Try this one instead.

32

12
number of vertices with more than two edges

If it is the number of ALL the lines, then the answer is 32. Same reasoning as before.

11.25

2^n duh, [1, 2, 4, 8, 16, 32, 64, 128 ... 2^n]

Ah fuck i get it now, 14.

...Or mabye it's 1,2,3,4,5,6,7,8...

...

Cannot assume a pattern from only 1 example.

Oh yeah! Thanks, user.

You could go by crossing non-corner lines, yielding 12. You could also go for amount of squares, yielding 4 small and one large for figure 1, or 9 + 4 + 1 = 14 for the large one. Pretty ambiguous puzzle, not worth wasting time on.

...

No, there's not enough information to uniquely discern a pattern. That just means there are many possible ``solutions''. Most are boring or convoluted, but OP's puzzle has at least one elegant `solution'. Your sequence, on the other hand, is truly boring.

4 boxes are area 5, each box therefore is area 5/4. Counting up 8 boxes is area of 40/4 or 10.

Or are you a potato?

>Autistic people often have difficulty 'reading' other people - recognising or understanding others' feelings and intentions

Its 21 why are you fucking retards saying things like 10 or 14

>not counting the four part blocks as 5
What

12 knots / intersections, where corners don't count

Correct.