Math Is Fun Forum

tony123 · 2009-03-11 09:43:16

find x

mathsyperson · 2009-03-11 10:08:02

tony123 · 2009-03-12 19:23:39

nice mathsyperson
but how

mathsyperson · 2009-03-13 00:31:09

Guesswork mainly.
x has to be in the form 88/k, since the LHS is x times an integer.
The LHS is also roughly x^4, which means that x is going to be roughly 3.

From there, just try different values of k to put into x/k until you find one that works.
(There is only one that does)

TheDude · 2009-03-13 02:48:24

Trial and error works really well here since we can narrow down possible answers to a workable amount.

First, we know that

is an integer, so x must be rational. Let a/b = x, GCD(a, b) = 1, then we have 88/x = 88b/a, which again is an integer. a must be a factor of 88, so a can be 1, 2, 4, 8, 11, 22, 44, 88. We can also see that 3 < x = a/b < 4. That let's us rule out 1, 2, 4, and 8 as possibilities for a.

We have now narrowed down our possible answers to these fractions:

11/3
22/7
44/13
88/23
88/25
88/27
88/29

You can then do trial and error on these 7 numbers.

Muggleton · 2009-03-15 09:02:26

TheDude wrote:

We can also see that 3 < x = a/b < 4.

How?

TheDude · 2009-03-15 15:12:15

3^4 = 81 and 4^4 = 256, so x must be between them.

Ricky · 2009-03-15 16:27:10

By using the quotient-remainder theorem, you can conclude:

It seems like you should also be able to get somewhere from here, but I can't see it.

Muggleton · 2009-03-21 12:34:19

I get it now. Thanks, TheDude.

Incidentally, are you all (apart from Ricky) only assuming that x is positive? Can't x be negative as well? After all, if you multiply a number four times, you do get a positive number. Ricky's method may allow for negative numbers but I don't really know what you can do with the method.

From all the numbers I've tried, I'm inclined to believe that there is no negative solution. I'm just wondering if you can definitely prove that no negative solutions exist.

mathsyperson · 2009-03-21 13:17:48

Interesting thought!
That function seems a lot jumpier in the negatives.
f(-3+ε) = 71
f(-3) = 81
f(-3-ε) = 111

So if we can show that's it's a decreasing function in the negatives then we've shown that there's no negative solution.

Math Is Fun Forum

#1 2009-03-11 09:43:16

x

#2 2009-03-11 10:08:02

Re: x

#3 2009-03-12 19:23:39

Re: x

#4 2009-03-13 00:31:09

Re: x

#5 2009-03-13 02:48:24

Re: x

#6 2009-03-15 09:02:26

Re: x

#7 2009-03-15 15:12:15

Re: x

#8 2009-03-15 16:27:10

Re: x

#9 2009-03-21 12:34:19

Re: x

#10 2009-03-21 13:17:48

Re: x

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