Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Looking for a way to generate random mathematical equations that equal an integer between 0 and 9, a way to rank the complexity of the equation rendered, and a method to note the mathematical concepts utilized within the equation.

My first guess is that a random integer between 0 and 9 would be selected. After that, your guess is much more likely to be better than mine.

EDIT: To be more clear, solutions should ONLY result in the following: 0,1,2,3,4,5,6,7,8,9.

share|improve this question
2  
What is the motivation behind this task? –  Yuval Filmus Jun 10 '11 at 20:54
    
Maybe he is making capture? –  Ilya Jun 10 '11 at 20:55
    
what kind of equations you are interested in? also, equations can not be equal to smth, are you talking about their solutions? –  Ilya Jun 10 '11 at 20:56
1  
@Gortaur: Do you have links to the meanings of capture and smth? –  blunders Jun 10 '11 at 21:09
    
@Yuval Filmus: My motivation is a secret. Is there anything about the question and the solution it seeks that is unclear to you? –  blunders Jun 10 '11 at 21:12

2 Answers 2

up vote 2 down vote accepted

I am still not sure what do you want, but I guess you are talking about roots. Ok, lets consider a class of linear equations. They have only two parameters: $ax+b = 0$. Choosing randomly $\hat{x} = 0,1,...,9$ you have a condition on $a$ and $b$: $a\hat{x}+b = 0$. for this class you can only randomly choose one of the variables (preferably, $b$ not to make $a$ equal to zero).

Example. $x = 2$, then $2a+b = 0$. You make $b = rand(0,100)$ and $a = -0.5 b$. then the solution will be $x=2$ as it was desired.

You can also consider another class - but note that you would like to have a unique solution. Say, for quadratic equation you may want to say "find the positive root of an equation". Then algorithm is the following: choose $\hat{x} = 0,1,...,9$ and $x'$ to be a negative number (preferably integer also). Then any equation of the form $a(x-\hat{x})(x-x') = 0$ has desired roots. For this equation you can choose independently $a$ (any) and $x'$ (negative). To make it more tough, open the brackets after the choice of all parameters.

Example. We choose again $\hat{x} = 2$ and $a = rand(0,12)$ and $x' = -rand(0,100)$. then we have an equation $ax^2 - (2+x')x + 2x' = 0$.

share|improve this answer
    
I'm assuming, in your first example, you mean for a solution $x = 2$, then $2ax + b = 0$...? –  amWhy Jun 10 '11 at 22:43
    
+1 @Gortaur: Thanks, stepped out, just got back. Answers looks fine, realized while I was out (looking at the question on my phone) that the question was flawed, though a good start, so I'm not going to edit the question since it's my error. Also, believe @amWhy is right, would you confirm that. Thanks! –  blunders Jun 11 '11 at 2:42
    
@blunders: I will not confirm it. This statement doesn't make sense to what I've written –  Ilya Jun 11 '11 at 13:34

Pick a positive integer n between 0 and 4 and a positive integer p between 1 and 5. Add r instances of (y+(-y)) to n+p, where r indicates a random positive integer, in other words (n+p)+(y_1+(-y)_1)+...+(y_r+(-y)_r). The complexity of the equation compared to other equations of the same type comes as the value equal to r. One could also select 4 positive integers between 0 and 2, or 5 positive integers two of which lie between 0 and 2 and three of which lie between 0 and 1, and many other possibilities exist. The partitions of 9 here work similarly.

share|improve this answer
    
+1 @Doug Spoonwood: complexity was meant to be the inclusion of concepts such as: addition, subtraction, division, multiplication, roots, fractions, powers, Decimals, percentages, etc. until as many concepts had been used that would not result in getting an answer that would not equal [0-9]. That said, your idea was creative, addressed two of the three parts of the question as far as I'm able to tell - though, to me, incremental redundant patterns produce volume, not complexity. As very possible I've misunderstood your answer, either way thank you! –  blunders Jun 11 '11 at 3:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.