[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math


View post   

File: 40 KB, 475x541, IMG_20170721_210216.jpg [View same] [iqdb] [saucenao] [google]
9053327 No.9053327 [Reply] [Original]

Show your working

>> No.9053328

2x2 square = 5 implies 3x3 square = 8

>> No.9053337

>>9053327
14

>> No.9053340

4x=5
X=5/4
3by3=y
9x=y
Y=13.5

>> No.9053342

>>9053327
14

>> No.9053344

>>9053327
12, the number of non-corner intersections

>> No.9053347

>>9053327
10
>Number of squares

>> No.9053350

>>9053327
there's not enough information to discern a pattern. if I told you "1,2, ___, ___, ___ ; fill in the blanks in this sequence" the appropriate response would be to look at me like I was a fucking idiot, not try to complete the task.

>> No.9053352

>>9053342
Going with this.

>> No.9053354

>>9053340
Without any more information this is literally the correct answer

>> No.9053364

[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]

>> No.9053366

>>9053347
>he thinks there are only 10 squares
there's 14

>> No.9053369

>>9053364
A correction:
Instead of [math]y_2 = 2^2 + 1[/math] I meant [math]y_2 = x_2 + 1[/math].

>> No.9053374

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

>> No.9053375

Number of squares or proportion.

>> No.9053376

>>9053374
Noice

>> No.9053377

>>9053327

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

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

>> No.9053386

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.

>> No.9053400

>>9053386
A nice excercise would be to put this solution into a closed form without recursion.

>> No.9053418

>>9053327
7

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

>> No.9053421

>>9053327
4 1x1 squares + 1 2x2 square = 5 squares
9 1x1 squares + 4 2x2 squares + 1 3x3 square = 15

>> No.9053424

>>9053327
Are individual squares of equal size? Because they don't appear to be.

>> No.9053431

12

>> No.9053441

>>9053421
>/sci/ still hasn't mastered addition
kek

>> No.9053443

>>9053327
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 =

>> No.9053446

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

>> No.9053467

>>9053441
oops 14 rather

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

>> No.9053506

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

>> No.9053518

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

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

>> No.9053522

>>9053518
No, wait!

I changed my mind.

It's (Number of small squares + 1)

4 + 1 = 5
9 + 1 = 10

>> No.9053525

>>9053518
>>9053522
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.9053527

>>9053518
>>9053522
>>9053525
No, no, no!

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

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

>> No.9053530

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.

>> No.9053543

kek at the fuggin underage sperg

>> No.9053564

>>9053327

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

>> No.9053571

>>9053400
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.

>> No.9053574

>>9053571
Right answer. >>9053421 had the same idea but fucked up when adding them together.

>> No.9053575

>>9053571
That's most likely correct but it's still a series, I meant the closed form of the summation

>> No.9053578

>>9053575
https://en.m.wikipedia.org/wiki/Square_pyramidal_number

>> No.9053580

>>9053578
Neat

>> No.9053609

>>9053327
4 + x = 5
x = 1

8 + 1 = 9

>> No.9053613

>>9053609
keep trying

>> No.9053616
File: 782 KB, 796x796, 1499564678928-fit.png [View same] [iqdb] [saucenao] [google]
9053616

72
Prove that I'm wrong faggot.

>> No.9053631

>>9053327

my first thought was just to solve for area

>> No.9053930

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.

>> No.9053932

>>9053327
14

>> No.9053941

>>9053327
If 4 boxes = 5 then 8 boxes should = 10

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

>> No.9053965

>>9053327
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!

>> No.9053968

>>9053965
ooops should be 9 divided by 4 = 2.25
so answer is actually 11.25
your very welcome!

>> No.9053974

5 intersections between lines up top (excluding the corners)

12 in the one below

>> No.9053977

>>9053340
i just did this and got 11.25 for Y

wut

>> No.9053981

>>9053941
and who gives a shit about 8 boxes?

>> No.9054001
File: 55 KB, 400x303, 1498579224639.jpg [View same] [iqdb] [saucenao] [google]
9054001

45/4 dumbasses.

>> No.9054123

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 ?

>> No.9054156

>>9054123
yeah, that's probably a finite field effect over mod 10

>> No.9054163

>>9054123
If you check, it's only the last digit in the prime that should have any affect on the last digit in the result

>> No.9054170

>>9053327
11,25

>> No.9054176

>>9054163
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?

>> No.9054180

1496569410
i forgot the zero

>> No.9054190

4= 5
9= 10

>> No.9054216

>>9053354
Its not. There is no reason to define the squares as x instead of, say points. The question doesnt have an answer.

>> No.9054302
File: 44 KB, 358x437, 1492663900233.jpg [View same] [iqdb] [saucenao] [google]
9054302

>>9053340
>45/4 = 13.5

>> No.9054487

The answer is 10.

>> No.9054497

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

>> No.9054553

>>9053327
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 :^)

>> No.9054559

>>9053327
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.
https://en.wikipedia.org/wiki/Planar_graph#Euler.27s_formula

>> No.9054592

>>9054559
Because the picture lacks information so everyone just goes with whatever shit.

>> No.9055182

bumping for answer

>> No.9055206

>>9053327
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

>> No.9055291

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

>> No.9056215

>>9053342
that

>> No.9056603

>>9053366
Number of smallest squares plus one
>>9053327
Question is retarded. You can make infinite patterns from a single data point

>> No.9056605

>>9056603
You can make infinite patterns from any number of data points

>> No.9056804

>>9056605
Don't you sass me Jimmy

>> No.9056832

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.

>> No.9056866

>Extrapolating from one datapoint
>>>/biz/

>> No.9056891
File: 228 KB, 1080x1920, 1486567772733.jpg [View same] [iqdb] [saucenao] [google]
9056891

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.

>> No.9057013
File: 17 KB, 853x714, solve.gif [View same] [iqdb] [saucenao] [google]
9057013

>>9053327
fuck your retarded retardings

>> No.9057059

>>9053350
"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.

>> No.9057067

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

>> No.9057079
File: 80 KB, 475x541, 1500667421447.jpg [View same] [iqdb] [saucenao] [google]
9057079

>>9053337
>>9053342
>>9053366
>>9053467
>>9053930
>>9053932
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.

>> No.9057118
File: 558 KB, 684x2101, 1459605310708.png [View same] [iqdb] [saucenao] [google]
9057118

>>9057079
32

>> No.9057124

>>9053327
12
number of vertices with more than two edges

>> No.9057135

>>9057079

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

>> No.9057147

>>9053327
11.25

>> No.9057156

>>9053350
2^n duh, [1, 2, 4, 8, 16, 32, 64, 128 ... 2^n]

>> No.9057159

Ah fuck i get it now, 14.

>> No.9057160

>>9057156

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

>> No.9057162
File: 26 KB, 677x378, Screenshot (15).png [View same] [iqdb] [saucenao] [google]
9057162

>> No.9057165

>>9053327
Cannot assume a pattern from only 1 example.

>> No.9057166

>>9057160
Oh yeah! Thanks, Anon.

>> No.9057207

>>9053327
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.9057208
File: 31 KB, 252x496, 1499049100978.jpg [View same] [iqdb] [saucenao] [google]
9057208

>>9057059

>> No.9057287

>>9053350

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.

>> No.9057302

>>9053981
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?

>> No.9057313
File: 266 KB, 1520x540, squarepuzzle.png [View same] [iqdb] [saucenao] [google]
9057313

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

>> No.9057314

>>9053327
Its 21 why are you fucking retards saying things like 10 or 14

>> No.9057318

>>9057313
>not counting the four part blocks as 5
What

>> No.9057352 [DELETED] 

>>9053327
let f(n) denote the number of subsquares in a square nxn grid.
we have:
f(1) = 1, f(2) = 5, ..., f(k) = 4*f(k-1) + 1, ...
solving this recurrence relation, we find:
f(k) = 4*f(k-1) + 1
= 4*4*f(k-2) + 4 + 1
= 4*4*4*f(k-3) + 4*4 + 4 + 1
= ...
= 4^(k-1) + sum_{i=0}^{k-2} 4^i
= 4^(k-1) + (4^(k-2))/3
= (4^k - 1)/3

>> No.9057364

>>9053327
12 knots / intersections, where corners don't count

>> No.9058222

>>9057135
Correct.

>> No.9058225

>>9057318
>dat deep level of irony here

>> No.9058231

5 squares in top diagram
4 1x1 and a 2x2

14 squares in bottom diagram
9 1x1, 4 2x2 and a 3x3

>> No.9058234

>>9053327
>>9053327
>>9053327

There are 4 squares which equal 5 ergo a square equals X.

4x=5
9x=y
9(5/4)=y
y=11.25

>> No.9058611

>>9057079
counted the number of vertices and multiplied by two
9 * 2 = 18
16 * 2 = 32

>> No.9058978

Jesus is the answer

>> No.9059184

>>9057314
you stupid

>> No.9060825

>>9053327
12.
1st diagram have 5 squares.
2nd diagram have 12 squares.

>> No.9060842

>>9053327
4 is 5
so 9 is 10 no I will not show my work thieves !
not that I'm the only one to arrive at this conclusion the problem is concussive both in practical and pattern .

>> No.9060888

understand that a whole is only equal to the sum of it's parts ! or have you midgets all be reduced to mindless pocket calculators ? The proof was in the plum pudding model all along *shakes head* The proof is in the pudding stop fucking blocking me for the internet you are sum of the most severely stupid "smart" people I've ever dealt with !

>> No.9060917

>>9057079
6(4-1) = 18
8(9-1) = 64

>> No.9060919

>>9053327
12.
First diagram contains 5 points with 3 or more intersections.
Second contains 12 of those same points.

>> No.9060932

you know it all depends on the answer the inquisitor was looking for ? but really not all things in science will follow a predictable pattern either not the one you may have expected such is nature .

>> No.9061003

>>9053327

1^2 + 2^2 = 5

1^2 + 2^2 + 3^2 = 14

>> No.9061005

>>9054176
>149656941
note the palindromic number

>> No.9061015

>>9056866
>one datapoint
Lrn2datum, Yank

>> No.9061289

>>9053327
5/4 = 1.25
1.25 * 9 = 11.25
I'm not sure if every reply in this thread is a joke.

>> No.9061311

-1/12

>> No.9061771

it is quite clearly the sum of small squares - 1 so answer is 8

>> No.9061801

>>9053327
not enough information, i can devise many systems that would produce a logical output

eg:
>how many small squares + 1
10
>number of vertices with more than 2 edges
12
>how many squares?
14
etc

>> No.9061804

13 = No of squares + nodes

>> No.9061809

>>9061804
those are internal nodes
if external included: 16 -9 = 7
for the small square 9 - 4 = 5

>> No.9062299

You're all gullible brainlets Jesus Christ