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#2 Re: Help Me ! » Powers of Two and Squares » Today 18:39:13

That is why I improved the NSolve command to only solve for Real numbers ( no imaginary).

You should see that only x = 59 and n = 6 are integers.

Since we know that there were only 4 possible solutions because there were only 4 pairs of divisors that when multiplied equal 615 we are almost done. Do you follow up to here?

There will be one more trick.

#3 Re: Help Me ! » Powers of Two and Squares » Today 18:12:47

I am sorry I inadvertently used k instead of n, let us go back to n and x as our variables. Check the whole calculation out now:

NSolve[{(2^n + x) == 615, (2^n - x) == 1}, {x, n}, Reals, WorkingPrecision -> 50]
NSolve[{(2^n + x) == 205, (2^n - x) == 3}, {x, n}, Reals, WorkingPrecision -> 50]
NSolve[{(2^n + x) == 123, (2^n - x) == 5}, {x, n}, Reals, WorkingPrecision -> 50]
NSolve[{(2^n + x) == 41, (2^n - x) == 15}, {x, n}, Reals, WorkingPrecision -> 50]

#4 Re: Help Me ! » Powers of Two and Squares » Today 18:00:54

Solve for k and x.

NSolve[{(2^k + x) == 615, (2^k - x) == 1}, {x, k}, Reals, WorkingPrecision -> 50]

What do you conclude?

#6 Re: Help Me ! » Powers of Two and Squares » Today 17:51:29

Starting with the first one:

It is reasonable to say,

#7 Re: Help Me ! » Powers of Two and Squares » Today 17:40:25

You got 4 pairs of numbers?

#9 Re: Help Me ! » Powers of Two and Squares » Today 17:30:39

You should have {1, 3, 5, 15, 41, 123, 205, 615}.

Now pair them off two at a time to make a product of 615. For instance 1 x 615 = 615.

#11 Re: Help Me ! » Powers of Two and Squares » Today 16:41:30

We will do it step by step and then it will get clear.

When you set the equation up as 2^n - x^2  = 615 you now factor 615

Divisors[615]

#12 Re: Help Me ! » Powers of Two and Squares » Today 15:45:22

Hi;

Try this piece of code:

Factor[2^(2 n) - x^2]

#13 Re: Help Me ! » Powers of Two and Squares » Today 12:37:17

It takes some math and some code to do that, M can do both.

#19 Re: Help Me ! » Powers of Two and Squares » Today 04:13:05

Hi;

That is the only one there is.

#20 Re: Help Me ! » Integrate ∫(1+4y^2)^(1/2) dy from 0 to 1 » Yesterday 15:15:18

Hi;

Is that e^(2y) in (y/4)+((e2y-e-2y)/16)?

#21 Re: Help Me ! » Problems » Yesterday 15:13:51

What were you trying to do with it?

#22 Re: Dark Discussions at Cafe Infinity » Are bobbym's neighbors zombies? » Yesterday 15:13:18

When people are so unaware of the world around them that they are not smart enough to just open the windows in cool weather...

#24 Re: Dark Discussions at Cafe Infinity » Are bobbym's neighbors zombies? » Yesterday 12:10:42

Nor would any sane person, the only thing that would is what I call a zombie.

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