## Numbers and Squares

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### Numbers and Squares

I don't know what the rules are for this puzzle. I don't know if it's an arithmetic operation, or something more complex. All I know is that it was given to Peruvian second graders, and my sister (who works with said kids) and I cannot figure it out.

sillybear25 wrote:But it's NPH, so it's creepy in the best possible way.

Shivahn wrote:I'm in your abstractions, burning your notions of masculinity.

### Re: Numbers and Squares

If there aren't any rules then I guess we can just make up our own. I'm not going to hide this with spoilers because it's not a great answer (but is there even a great answer?), but it's an answer.

Add up all the numbers for a square:

Square 1: 8 + 4 + 9 + 2 + 6 = 29

Square 2: 2 + 17 + 6 + 10 + 21 = 56

The difference between the first two sums is 27.

Square 3: 12 + 5 + 8 + 8 = 33

In order to get a difference between the sums from the 2nd and 3rd square, the numbers have to add up to 83, so the missing number is 50.

Like I said, not a great answer, but if there are no rules, then this answer is as good as any and better than many.

Add up all the numbers for a square:

Square 1: 8 + 4 + 9 + 2 + 6 = 29

Square 2: 2 + 17 + 6 + 10 + 21 = 56

The difference between the first two sums is 27.

Square 3: 12 + 5 + 8 + 8 = 33

In order to get a difference between the sums from the 2nd and 3rd square, the numbers have to add up to 83, so the missing number is 50.

Like I said, not a great answer, but if there are no rules, then this answer is as good as any and better than many.

### Re: Numbers and Squares

Since we're giving crappy answers...

You divide the middle number by 3, and add it to the bottom right number to get the top right number. Therefore the 3rd middle number is -9.

OBVIOUSLY.

You divide the middle number by 3, and add it to the bottom right number to get the top right number. Therefore the 3rd middle number is -9.

OBVIOUSLY.

-Adam

### Re: Numbers and Squares

I was assuming that the answer was of the form f(c1,c2,c3,c4) = m, where c1-c4 are the numbers on the corners and m is the number in the middle. I can't find a convenient, simple function that actually works, though. Obviously if you get complex enough, there are going to be an infiniute number of possible functions.

### Re: Numbers and Squares

^Above posts: When mathematicians try to solve simple puzzles :p

### Re: Numbers and Squares

Adacore wrote:I was assuming that the answer was of the form f(c1,c2,c3,c4) = m, where c1-c4 are the numbers on the corners and m is the number in the middle. I can't find a convenient, simple function that actually works, though. Obviously if you get complex enough, there are going to be an infiniute number of possible functions.

That was the theory I was going on. It's not terribly difficult to find solutions involving two or three numbers, but none I've seen work with all 4.

sillybear25 wrote:But it's NPH, so it's creepy in the best possible way.

Shivahn wrote:I'm in your abstractions, burning your notions of masculinity.

### Re: Numbers and Squares

The EGE wrote:Adacore wrote:I was assuming that the answer was of the form f(c1,c2,c3,c4) = m, where c1-c4 are the numbers on the corners and m is the number in the middle. I can't find a convenient, simple function that actually works, though. Obviously if you get complex enough, there are going to be an infiniute number of possible functions.

That was the theory I was going on. It's not terribly difficult to find solutions involving two or three numbers, but none I've seen work with all 4.

A function only using 2 of the 4 variables is still a function in 4 variables. c.f y=1, which still makes a perfectly good graph of a function.

addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

### Re: Numbers and Squares

mike-l wrote:A function only using 2 of the 4 variables is still a function in 4 variables. c.f y=1, which still makes a perfectly good graph of a function.

It would be a pretty shitty puzzle if half the numbers weren't relevant to the solution.

I made a program that attempted to combine the 4 outer numbers with simple arithmetic (+ - * /) in every possible way, but out of over 9000 functions, none worked for both of the first two squares.

### Re: Numbers and Squares

call the numbers a, b, c, and d counting clockwise from the top left. Then, center= (3/10)(2d-a)(b-c) works and gives -3.6 for the missing number.

### Re: Numbers and Squares

Goplat wrote:It would be a pretty shitty puzzle if half the numbers weren't relevant to the solution.

Red herrings FTW! I think we can all agree that this is a pretty shitty puzzle regardless of how many numbers are relevant to the solution.

-Adam

### Re: Numbers and Squares

Wild speculation:

**Spoiler:**

### Re: Numbers and Squares

Here's an answer that squeezes the facts until they fit, and produces the same end result as GeorgeH.

**Spoiler:**

### Re: Numbers and Squares

I did wonder if it might be a word puzzle (since the numbers are all between 1 and 26), but nothing falls out easily, so I don't think there's enough data to get anything sensible out of that idea.

### Re: Numbers and Squares

If it is a word problem, perhaps it is in Spanish? These are Peruvian second graders, after all.

This may be one of those 'pee in the pipe to get the ping pong ball out' type problems that my adult brain can't process. Tonight, on 'Are You Smarter Than A Peruvian Second Grader,' I don't walk away with a million nuevo sol.

This may be one of those 'pee in the pipe to get the ping pong ball out' type problems that my adult brain can't process. Tonight, on 'Are You Smarter Than A Peruvian Second Grader,' I don't walk away with a million nuevo sol.

Is 'no' your answer to this question?

### Re: Numbers and Squares

I got one...

Take the sum of the outside numbers the boxes (OBox):

OBox A = 23

OBox B = 35

OBox C = 33

take the change between the numbers:

OBox A +15 = OBox B

OBox B -2 = OBox C

take the change and add it to the inside box number (IBox) to get the next (IBox) number:

IBox A (6) +15 = IBox B (21)

therefore:

IBox B (21) -2 = IBox C (40)

The Answer is 40

Take the sum of the outside numbers the boxes (OBox):

OBox A = 23

OBox B = 35

OBox C = 33

take the change between the numbers:

OBox A +15 = OBox B

OBox B -2 = OBox C

take the change and add it to the inside box number (IBox) to get the next (IBox) number:

IBox A (6) +15 = IBox B (21)

therefore:

IBox B (21) -2 = IBox C (40)

The Answer is 40

Disclaimer: I am not a scientist.

### Re: Numbers and Squares

Change between OBox A and OBox B is 12, not 15.

sillybear25 wrote:But it's NPH, so it's creepy in the best possible way.

Shivahn wrote:I'm in your abstractions, burning your notions of masculinity.

- TheDancingFox
**Posts:**13**Joined:**Sat Aug 13, 2011 5:58 am UTC

### Re: Numbers and Squares

Yeeeeah, I'm pretty sure all of the answers so far are correct? I would presume that if there's a valid puzzle there's an answer properly including all four numbers, and I wouldn't expect it to be super complex given the context. Just makes the whole thing weird. Honestly, the psychology theory seems the most plausible so far (not necessarily the answer it gives, but just that that's what this thing is).

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