[EventHorizon]

(1) This one is taken from a math puzzle website (which I cannot name here due to mods removing offsite puzzle references).

What are all possible pairs of integer values that satisfy the following equation? And can you prove there are no more?

x^y = y^x

(2) I couldn't sleep one night, and noticed the clock's time had an interesting property. Let the clock's time be x:yz am. x is a factor of yz (not y*z, but simply putting their digits together), z is a factor of xy, and y is a factor of xz. Not only this, but yz/x = xy/z. Additionally, x, y, and z are all unique. What are x, y, and z?

I'm not sure if there is more than one solution to this one....and if there is I can add another constraint.

[Guest]

(2) I couldn't sleep one night, and noticed the clock's time had an interesting property. Let the clock's time be x:yz am. x is a factor of yz (not y*z, but simply putting their digits together), z is a factor of xy, and y is a factor of xz. Not only this, but yz/x = xy/z. Additionally, x, y, and z are all unique. What are x, y, and z?

I'm not sure if there is more than one solution to this one....and if there is I can add another constraint.

3:24 am

3 is a factor or 24

2 is a factor of 34

4 is a factor of 32

24/3=32/4

[EventHorizon]

3:24 am

3 is a factor or 24

2 is a factor of 34

4 is a factor of 32

24/3=32/4

Yup, that's what time it was. Good job.

[Guest]

(1) This one is taken from a math puzzle website (which I cannot name here due to mods removing offsite puzzle references).

What are all possible pairs of integer values that satisfy the following equation? And can you prove there are no more?

x^y = y^x

2^4 = 4^2 = 16

[Guest]

and of course, all numbers where x=y

[Guest]

For the first question: x^y = y^x,

Actually, it is a pretty easy and well creative question. Because the question just tell us to make it works, we can easily pick a sepcial number: 0. When 0 insert to both x and y, the equation will be form.

Under this path, we can make x=y. That way, no matter what numbers you are going to pick, it is always equal.

Thank you for your attention

FallingLeave

[Guest]

2^4 = 4^2 = 16

(-4)^(-2) = (-2)^(-4)

The negative works as well.

[Yoruichi-san]

You can always write y as y=x^a (where a is not necessarily an integer) by taking a = log(base x) of y.

So we can rewrite x^y=y^x as:

x^(x^a)=(x^a)^x

Using the fact that (A^B)^C =A^(B*C), the right side can be rewritten, and then we have:

x^(x^a)=x^(a*x)

In order for this to be true, then we have to have (x^a)=(a*x), which is true for a=1 (making x=y), or a=2 and x=+/-2.

[Guest]

The equation can be rewritten as ln(x)/x = ln(y)/y, since a quick check demonstrates that neither can be 0.

x=y is a trivial solution, so now we look for other solutions.

Since the equation is symetric, if we graphed f(n)=ln(n)/n we would want to find two different n values that generate the same f(n) value.

f'(n) = (1-ln(n))/n^2, thus when n<e, f(n) is increasing, and when n>e, f(n) is decreasing. Since the distinct n values must generate the same f(n) value, we must have a pair of n's such that one is greater than e and the other less than e.

Noting that the problem limits us to integers, all we need to check is n=1 and n=2 and see if the corresponding value is also an integer.

A quick check shows that 1 has no solution, and we get the pair (2,4).

[Guest]

Hello everyone this is my first visit as a registered member. I love this site

(1) This one is taken from a math puzzle website (which I cannot name here due to mods removing offsite puzzle references).

What are all possible pairs of integer values that satisfy the following equation? And can you prove there are no more?

x^y = y^x

(2) I couldn't sleep one night, and noticed the clock's time had an interesting property. Let the clock's time be x:yz am. x is a factor of yz (not y*z, but simply putting their digits together), z is a factor of xy, and y is a factor of xz. Not only this, but yz/x = xy/z. Additionally, x, y, and z are all unique. What are x, y, and z?

I'm not sure if there is more than one solution to this one....and if there is I can add another constraint.

6:48 am if you had a very bad night

[Guest]

(1) This one is taken from a math puzzle website (which I cannot name here due to mods removing offsite puzzle references).

What are all possible pairs of integer values that satisfy the following equation? And can you prove there are no more?

x^y = y^x

2^4=4^2 works