The Hardest Maths Questions You'll Ever Try

  • Thread starter Doomotron
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Doomotron

Doomotron

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Doomotron
Welcome to The Hardest Maths Question You'll Ever Try, my thread where we can challenge each other with the most confusing thing in the universe, maths. You can post a question and we'll try to answer it.

P. S You can see the old OP below, when this thread had a different purpose.
This is a thing I made up using my will to confuse the internet and my maths 'skills'. Every week I'll set a question to be answered.

First you'll have to understand how this works. The numbers turn into letters, so A is 1, B is 2, and so on. After going passed Z, you start with double letters: AA, AB, AC, and when you get to AZ, you start again with BA, and so on. Knowing this, you can answer all the questions. I will set the first question on 24/3/16 or 3/24/16 if you live somewhere else.
 
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The hardest math questions I'll ever try? Good luck, I've used the Navier Stokes equations:

2630552_orig.jpg


...and that wasn't even the worst of it.
 
A general math puzzle thread would be cool. @Doomotron , is it ok to post our own puzzles here? If not, I'll start another one if there's enough interest.
 
This might be intriguing for me too - although I'm not very good at math and you can't really expect me to give clever answers if I ever try this stuff.
 
AspecBob
A general math puzzle thread would be cool. @Doomotron , is it ok to post our own puzzles here? If not, I'll start another one if there's enough interest.
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Don't worry. You can post them here.

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Here is my first question:

If Z-B=X, would A+Z=X be correct?
 
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Just to say, if you are going to answer more than one question in a post, please use Quote to show it. As you can't add quotes in Edit Mode, just put @<User Name>.
 
Doomotron
Don't worry. You can post them here.

_____________________________________________________________________________________________________

Here is my first question:

If Z-A=X, would A+Z=X be correct?
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What the hell is that? Its all wrong by algebra standards.
 
Doomotron
Please read the first post again. My questions do not involve algebra, the include a letter/number mix. A is 1, B is 2,...
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Either it's following algebra, or it's gibberish. If the alphabet is corresponding to counting numbers, your very first question is invalid:

26 - 1 = 24 / 1 + 26 = 24

As an aside: Why do you insist on making yourself the focus of every new thread? You've got your own name listed 6 (six!) times in your signature.
 
BobK
First of all Z-A does not equal X, it equals Y. Z - B would equal X.

As for the question, no.
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That's what makes the question hard! The answer is yes. Follow me here. ;)

Doomotron
First you'll have to understand how this works. The numbers turn into letters, so A is 1, B is 2, and so on. After going passed Z, you start with double letters: AA, AB, AC, and when you get to AZ, you start again with BA, and so on. Knowing this, you can answer all the questions. I will set the first question on 24/3/16 or 3/24/16 if you live somewhere else.
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So we know that Z is 26, A is 1, and X is 24. We're asked to take the following as premises:

A=1
Z=26
X=24
Z-A=X

Our first step of logical evaluation is then to substitute the number for the letters

26-1=24

Consolidating, that's:

25=24

Here, we've shown a contradiction. That A equals not A. According to the principle of explosion, anything follows a contradiction. So we can conclude A+Z=X, or anything we'd like from these contradictory premises. That's not as hard as the navier stokes equations though.
 
@Doomotron How old are you?

I was expecting at least some kind of algebra or calculus, which could be the hardest kind of math for the average person around the age of 16.

I'll leave an easy problem here, in case anyone wants to solve it.

We have these three equations:
5x +3y +2z =11
-3x +2x -7z =-26
8x -y +z =3

Solve for x, y and z

Use any method you want (Though there's no actual accomplishment by doing it by trial and error, you can do it like that if you want)
 
According to my calculations, this problem is impossible to solve. To be precise, the answer to this equation is I don't know.
 
My problem or his problem.

Because his problem is impossible to solve, I took mine from a test I did last semester and is totally possible to solve.
 
I suggest that posted solutions be put in a spoiler tag in case there are people who are still working on the problem.

Here's mine:

Using only the four basic operations (+, -, *, /), the numbers 1, 5, 6 and 7, and as many parentheses as you want, make 21. (Each of the four numbers must be used once.)
 
AspecBob
Here's mine:
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I think a change in wording is needed. Is it meant to be only one use for each number? As is, with no limit on how many times we can use them, the easiest solution is simply 7+6+5+1+1+1.
 

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