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posted 8 years ago

Hello guys...!

It's very funny because my little brother asked me this question;

What are the possibilities of a, b, and c ?

Could any body help me,

1st) What topic in math is this that I should look upon?

2nd) Step by step how to solve it (the way)...?

3rd)... your kindness....

Somebody outthere give me short explaination meanwhile, I need to erase my stupidity because of not aware of this calculation in my head. Anyway here is the words

But here, my concern is... why 2^9 could be**2^4 + 2^3 + 2^2** instead of **2^4 x 2^3 x 2^2**

for that solving case?

It's very funny because my little brother asked me this question;

What are the possibilities of a, b, and c ?

Could any body help me,

1st) What topic in math is this that I should look upon?

2nd) Step by step how to solve it (the way)...?

3rd)... your kindness....

Somebody outthere give me short explaination meanwhile, I need to erase my stupidity because of not aware of this calculation in my head. Anyway here is the words

But here, my concern is... why 2^9 could be

for that solving case?

posted 8 years ago

If it is really a math question and not a trick question, then there is not enough information to solve for a, b and c.

For three unknowns, you need three distinct equations to solve the problem.

For three unknowns, you need three distinct equations to solve the problem.

posted 8 years ago

I suspect the question is asking for (positive?) integer solutions. Which will at least give you a finite number of possibilites.

Bear Bibeault wrote:If it is really a math question and not a trick question, then there is not enough information to solve for a, b and c.

For three unknowns, you need three distinct equations to solve the problem.

I suspect the question is asking for (positive?) integer solutions. Which will at least give you a finite number of possibilites.

posted 8 years ago

Solving uniquely for three variables takes three linear equations, but abc = 1536 is not linear. I remember one branch of math where you could optimize one formula based on an insufficient number of linear equations. I believe it was called linear programing, though it had nothing to do with computers.

In any case, that's not what you're doing here. You're looking for a fast way to find the integer solution to these equations. Factoring 1536 into 3 * 2^9 is a good way to start because it gives you all the factors that will make up a, b, and c. It's a bit hard to explain if that doesn't click for you. 2^9 of course doesn't equal 2^4 + 2^3 + 2^2, but you will build up a, b, and c by multiplying one 3 with nine 2s. That gives you a quick way to limit your search for a solution.

When I just did it on paper, I started with a = 6 (3x2), and b = 16 (2x2x2x2), which left four of our nines 2s so c was forced to be 16. (I'm finding it difficult to explain what I mean by "left four of nine 2s". I think you'll either get it, or you won't.) Anyway, that meant the sum of a,b, and c was 38, almost right but not quite. So I thought to bring one of those 2s over to a, making it 12, and taking it away from b, making it 8. That kept c at 16. So, I got the answer in two guesses.

The technique of making a guess, and comparing its results to the true solution, and then adjusting, is called*regula falsi*. I don't know a name for the rest of what I did. Maybe we can call it JFM?

In any case, that's not what you're doing here. You're looking for a fast way to find the integer solution to these equations. Factoring 1536 into 3 * 2^9 is a good way to start because it gives you all the factors that will make up a, b, and c. It's a bit hard to explain if that doesn't click for you. 2^9 of course doesn't equal 2^4 + 2^3 + 2^2, but you will build up a, b, and c by multiplying one 3 with nine 2s. That gives you a quick way to limit your search for a solution.

When I just did it on paper, I started with a = 6 (3x2), and b = 16 (2x2x2x2), which left four of our nines 2s so c was forced to be 16. (I'm finding it difficult to explain what I mean by "left four of nine 2s". I think you'll either get it, or you won't.) Anyway, that meant the sum of a,b, and c was 38, almost right but not quite. So I thought to bring one of those 2s over to a, making it 12, and taking it away from b, making it 8. That kept c at 16. So, I got the answer in two guesses.

The technique of making a guess, and comparing its results to the true solution, and then adjusting, is called

posted 8 years ago

Why is abc = 1536 non-linear? I assume it means a * b * c = 1536.

each variable is a first power, and the constant is 1 in each case...

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.

each variable is a first power, and the constant is 1 in each case...

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Matthew Brown

Bartender

Posts: 4568

9

posted 8 years ago

Combinatorics? The brute force approach involves generating all possible combinations of three factors, and then checking the second condition.

Here's one pattern you could follow:

Of course, you can instantly rule out any combinations with 2^6 and above, or 3x2^4 and above, as they instantly exceed the target.

(All assuming we're talking positive integers, of course)

Here's one pattern you could follow:

Of course, you can instantly rule out any combinations with 2^6 and above, or 3x2^4 and above, as they instantly exceed the target.

(All assuming we're talking positive integers, of course)

Matthew Brown

Bartender

Posts: 4568

9

posted 8 years ago

Because you're multiplying variables together. It doesn't satisfy the "single variable" part of that definition.

fred rosenberger wrote:Why is abc = 1536 non-linear? I assume it means a * b * c = 1536.

Because you're multiplying variables together. It doesn't satisfy the "single variable" part of that definition.

posted 8 years ago

Factorize and permutate...

posted 8 years ago

Taking the brute force approach and to have some fun with Haskell:

which returned [(8,12,16),(8,16,12),(12,16,8)]

which returned [(8,12,16),(8,16,12),(12,16,8)]

- Marimuthu Madasamy

posted 8 years ago

Very nice.

If we allow negative numbers (and why not?), there's another set of solutions: (-8,-4,48) and its permutations.

If we allow negative numbers (and why not?), there's another set of solutions: (-8,-4,48) and its permutations.

posted 8 years ago

duh...sorry.
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Matthew Brown wrote:

fred rosenberger wrote:Why is abc = 1536 non-linear? I assume it means a * b * c = 1536.

Because you're multiplying variables together. It doesn't satisfy the "single variable" part of that definition.

duh...sorry.

J. Insi

Ranch Hand

Posts: 90

posted 8 years ago

waaah, I never calculate in this manner.... guys, are you all taking math class right now?

Marimuthu Madasamy wrote:Taking the brute force approach and to have some fun with Haskell:

which returned [(8,12,16),(8,16,12),(12,16,8)]

waaah, I never calculate in this manner.... guys, are you all taking math class right now?

Mike Simmons

Ranch Foreman

Posts: 3268

20

posted 8 years ago

Not really. It's a program in the Haskell programming language. Personally I was learning about it as part of reading Seven Languages in Seven Weeks, which we had as a reading group at work. (By the way, if you do this: I recommend taking more than seven weeks. Maybe fourteen weeks. You can learn a lot if you work through all the problems, but it can take a lot of time.) I don't know if I will ever program in Haskell professionally, but it's a really good way to stretch your brain.

posted 8 years ago

Though Mathematical skills would help in programming, the above code is straightforward, taking the brute force approach. Further I also have been learning Haskell for one month and really love the language. Learning Haskell would definitely improve your programming skills and your way of thinking in programming.

- Marimuthu Madasamy

J. Insi

Ranch Hand

Posts: 90

posted 8 years ago

waw.... i've never involved in what that haskell thing....

great Mike, keep it up! is it difficult than our programming language or ....

Mike Simmons wrote:Not really. It's a program in the Haskell programming language. Personally I was learning about it as part of reading Seven Languages in Seven Weeks, which we had as a reading group at work. (By the way, if you do this: I recommend taking more than seven weeks. Maybe fourteen weeks. You can learn a lot if you work through all the problems, but it can take a lot of time.) I don't know if I will ever program in Haskell professionally, but it's a really good way to stretch your brain.

waw.... i've never involved in what that haskell thing....

great Mike, keep it up! is it difficult than our programming language or ....

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