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        riddles >> medium >> One real solution
        
(Message started by: rugga on Oct 5^th, 2002, 1:03am)

Title: One real solution
Post by rugga on Oct 5^th, 2002, 1:03am

Mathematics departments at some south-western universities received
Quote:

Mr. H.N.’s sly letters asking for the one real solution x of the following two equations:

18 = ((1 + x)¹⁸/x)
17 = ((1 + x)¹⁷/x)

Professor A.S. at one of those departments sent Mr. H.N. the following brief solution:

18/17 = ((1 + x)¹⁸/x) / ((1 + x)¹⁷/x) = 1 + x , so x = 1/17

Are there any other real solutions x? Why?

There's a clever deception going on here. Let me just start by pointing out that it's easy to verify by calculator that x = 1/17 is not actually a solution to either equation.

I guess the flaw in the "solution" is in combining two unrelated equations of the same unknown. It's as if you asked what the common solution is for
x³ = -8 and
x² = 1.
Using the logic above, we can divide one equation by the other to get:
x = -8. QED :-/

As far as "other" solutions go, here's a couple hints:
1) Try to solve the first equation by itself.

2) If stuck with hint #1, take the derivative of the function f(x) = (1 + x)¹⁸ / x
to see what the minima and maxima are.

One thing that still puzzles me is who are H.N. and A.S.? Am I missing the joke?

Title: Re: One real solution
Post by NickH on Oct 5^th, 2002, 5:24am

Nice one! I'm not sure exactly what the joke is. Perhaps it is at the expense of math students!

Here's another way to see that the first equation has no real solution.

Assume there is a positive solution.
Then (1 + x)^18 = 18x, and so:
1 + 18x + 153x^2 + ... = 18x.
Subtracting 18x from both sides, this is clearly impossible for positive real x!

Then, x < 0 => (1+x)^18 / x < 0, so there is no negative solution.

Similarly, there is no positive solution to the second equation. Neither is there a solution with -1 <= x < 0. However, there must be at least one real solution because (1 + x)^17 - 17x = 0 is a polynomial of odd degree.

Let u = 1+x, then f(u) = u^17 - 17u + 17 = 0.
So fdash(u) = 17(u^16 - 1).
There is a minimum at (1,1) and a maximum at (-1,33).

Therefore there is one real solution, with x < -2.

Nick

Title: Re: One real solution
Post by kenk on Oct 10^th, 2002, 12:28pm

This seems to me to be a trick question. Let me reword a bit.

Consider the two functions:

f(x) = ((1+x)^18/x)/18
g(x) - ((1+x)^17/x)/17

Find a real x such that:

1) f(x) = g(x) (both not infinity)
and
2) f(x) = 1

OK, let's tackle part 1. Clearly 1) is satisfied iff f(x) = g(x) = 0 or f(x)/g(x) = 1 (and g(x) != 0).

f(x) and g(x) are both zero for x = -1, but for no other values.
f(x)/g(x) = (1+x) * 17/18. If f(x)/g(x) = 1, then 1+x = 18/17, so x = 1/17.

Now, we need merely examine these two values to see where f(x) = 1.

f(-1) = 0. No good.
f(1/17) = (18/17)^18*17/18 = (18/17)^17 > 1.

Thus, there are NO real solutions to those equations.

Title: Re: One real solution
Post by Chronos on Oct 14^th, 2002, 7:00pm

Or, to put it another way: If there is one simultaneous real solution to both equations, then x = 1/17.

Title: Re: One real solution
Post by william wu on Dec 8^th, 2002, 3:34pm

It's all a true story. H.N. and A.S. are initials of real math professors. Unfortunately I don't know their full names :)