Rules:

NOpe

That would be telling ;)

Dam I thought this was most reasonable
Guess there's still a chance

I will give you a few more minutes

Hint:

Maybe that is not a Z ...

It has to be horizontal? Or i could look at it vertical?

00 is identical to 0= ? Meh i'm running out of ideas

00≠0-1

If not I give up

Even the self proclaimed 'most intelligent being on earth' doesn't understand it

A hint:
That's a 5, not a Z, don't be stupid

0050-1

00

Not op, but what? U made the 5 by squishing the bottom of the 5 into the equals sign? That still doesn't use the hint

The answer was:

COS 0 = 1

Here is another one

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Roman5 = 5

Fucking basic trigo and i was shitted on

Wrong.

A roman 5 is a "V", not a "IIIII"

What I meant is the 5 with the IIII and the slash... my bad

That is correct.

Next one

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(X^3 1 = 1)

How did you even get there, moving one?

The righter C. I just moved the left matchstick

I still dont see it, but anyway, no, that is not the right answer.

What does that even mean. How does it equal 12?

Its called implicit multiplication

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I got it. Both fractions are multiplying, though there's no sign between them

I've never seen it written like that with no sign in between. At least here in Europe I've always used either the multiplication sign or parentheses

Im European too. I have seen it plenty of times.

After all, if you read

2x + 5

You know it means

2*x + 5

I hope this thread doesnt die. My phone will be dead in a bit

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Yes, but that's because you're multiplying a number by a variable. When you multiply a number by another number I've always seen and used the either the multiplication sign (which is a cross here) or parentheses

LOL!!!

No, Those do not count as parenthesis :D

But very creative idea :D

But how does "X" resolve? If you have an unresovled variabled, you cannot check if both sides match...

You are not simply multiplying 2 numbers. You are multiplying 2 fractions. In those cases, the implicit multiplication works.

Sorry, but the given solution is the right one.

You have to twist a little your imagination, if you wanna solve this puzzles. The ones coming are even weirder :)

I think this has something to do with x/x=1. But then again for that we need to know that x≠0, unless that'd be assumed by the type of question this is.

You have to cancel the variables to reach an equality, unless the x's mean the multiplying sign instead of the variable x

HAHAHA had to try

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Ok, here is the solution:

CX / XI = 10

Roman numbers

110 / 11 = 10

Next one.

Hint: Things might not be exactly what they seem...

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Can you tilt the sticks to make a multiplication sign?
You could make (1*2)(4/2)=4

To do that, you have to move 3 sticks

move the minus sign and the fraction line to make the 4 a 9 so it becomes a dot product between two matrices (1 2)(5 2) = 9

Genius

Correct!!! Very impressive

Another one

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Where the fuck did this thread come from? Intelligence and creativity on b? Gg OP.

move the horizontal matchstick in the 7 to make the second - a +
(4-1)(4+2)^1 = 18

Instead of 7 do a 1 and move that to make a 4+2 instead of 4-2

(4 - 1 X 4 + 2)^1 = 18

First thought too but I dont think a slanting 1 counts?

I'll take my victory in this one as I had not seen your answer yet

But a / isn't a 1

Use your imagination and instead of positioning the match like / do it like |

But then you move two sticks retard

(4-1x4+2) = 2

move the top right stick from the 8 and angle it so the 7 looks like a 4: (4 - 1 x 4 - 2)^4 = 16

nice, but if this works, so do these

Wrong. You need to more 2 matches to make a 1

This is not the right solution

But you have parentheses between the numbers

I’m stuck

My only thought is 2-> but that yields

^7 = 18

Alright, since we're stuck at this, let's go to the real math questions, starting easy:
Prove that every odd number can be represented as the difference of two perfect squares

difference of two consecutive perfect squares is (x+1)^2 - x^2 which expands to x^2+2x+1-x^2 = 2x+1. Every odd number can be represented by 2x+1

X^2-y^2 = z where x is perfect square 1, ye is perfect square 2 and z is odd number

Very well. Now, a harder one:
Prove that if you add 1 to the product of four consecutive integers you'll always get a perfect square

Can we get back to math problems or matchstick problems because stuff like this exposes the googlers, which ruin this whole thread

I'm a different user than the OP posting those match problems. Since we were stuck at the match problems, I figured I would add a bit of variation

Here is the solution

I have more, but I have no more time today. Sorry, anons. Another day.

Kisses

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*correction: four consecutive Positive integers

I think this is too advanced for me. I haven't learned matrices yet, and I don't know the meaning of that T

In my opinion - correct. I think this same.

Thats all it means

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multiplying 4 consecutive positive integers plus 1 gives:
x*(x+1)*(x+2)*(x+3)+1
when expanded, this will give
x^4 + 6x^3 + 11x^2 + 6x +1
you can factor this to:
(x^2 + 3x + 1)^2
which is always a perfect square

you were right the first time this one will give 14

youve just moved two of them no? the minus and the bottom half of the 7

Here is a fast one. I couldnt hold it :)

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How did you factor that? I don't think you can do it by hand

Sorry, i'm stupid

I noticed the coefficients' symmetry and assumed it was a trinomial squared based on the original question and since the first and last coefficients are 1 all you need to do is find the middle one

Idk if this is correct

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Correct

100 = 10^2 which is a big square :)

Now I have to leave. Good night, anons

Goodnight user thanks for actually giving my head some exercises

Ahahahha, That's good took a few seconds lol

This is a more elegant way
x(x+1)(x+2)(x+3)+1=
=(x^2+3x+2)(x^2+3x)+1=
=(x^2+3x+1+1)(x^2+3x)+1=
=(x^2+3x+1)(x^2+3x)+x^2+3x+1=
=(x^2+3x+1)(x^2+3x+1)=
=(x^2+3x+1)^2

You know what, im out of the door, but I will leave you with one of the riddles i DO NOT KNOW the solution of.

Good luck, and this time, really, bye

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Let p be a prime number greater or equal to 5.
Prove that for any value of p, p^2-1 is a multiple of 24