Find the lowest n from N that satisfies:

[Eliva]

Homework Statement

Find the lowest number n from N that satisfies:

Homework Equations

http://img72.imageshack.us/my.php?image=zadacaqw8.jpg
Please see attached
if k is any number from N

The Attempt at a Solution

I have absolutely no idea where to start. :(

 

[tiny-tim]

Hi Eliva!

If the attachment is just the equation that you have to prove, then please type it out for us.

It's not fair on the moderators to make them do the work of checking your attachment when it would be far less work for you to type it out.

 

[HallsofIvy]

the attachment says
a/2= k₁²
a/3= k₂²
a/5= k₃²

Since his statement of the problem say simply "Find the smalllest number from N that satisfies" and ends there, this makes no sense at all.

 

[Eliva]

Hi Eliva!

If the attachment is just the equation that you have to prove, then please type it out for us.

It's not fair on the moderators to make them do the work of checking your attachment when it would be far less work for you to type it out.

Yeah, well, I tried ... the Latex reference was not displaying the equations right, that's why i did it in Word and uploaded an image. I am sorry if that put u into trouble, but the code was not working right, all it gave me in the preview was a few red lines. So again, I am sorry.

the attachment says
a/2= k₁²
a/3= k₂²
a/5= k₃²

Since his statement of the problem say simply "Find the smalllest number from N that satisfies" and ends there, this makes no sense at all.

Yes, I might have not explained it quite right. The thing is that u acctually need to find the number a, if all those three statements are correct, which at the same time is suppose to be the lowest (smallest) natural number. As in my knowledge it might have smth to do with finding the GCD of these numbers, or maybe use the Euclidean algorithm, but I am not sure exactly how. k₁, k₂, and k₃ are merely constants, they can be any number, it only means that for instance, in the first statement a/2 exists as a full square from some number K. Please have in mind that all three statements must be correct for the number a.

 

[HallsofIvy]

So a= 2k<sub>1²= 3k2²= 5k3² for some numbers k1, k2, k3?

If the k's and a are required to be integers, there is no such number: the first equation requires that a= sqrt(2) k1 and, since sqrt(2) is not rational, that is impossible.

I the k's and a are allowed to be general real numbers, there is no smallest such number- just take the k's to be arbitrarily close to 0.</sub>

 

[tiny-tim]

the attachment says
a/2= k₁²
a/3= k₂²
a/5= k₃²

Hi HallsofIvy!

Well, I can see the .jpg now, but it actually says
a/2= k₁²
a/3= k₂³
a/5= k₃⁵

Hi Eliva!

So you mean the question is "In the natural numbers, N, find the lowest such that half of it is a perfect square, one-third is a perfect cube, and one fifth is a perfect fifth power"?

Hint: Start by writing out some equations which combine more than one of the three basic equations (and call them p q and r rather than k1 k2 and k3 … it's easier to type ).

What do you get?

 

[Eliva]

So you mean the question is "In the natural numbers, N, find the lowest such that half of it is a perfect square, one-third is a perfect cube, and one fifth is a perfect fifth power"?

Yessss! that's exactly what I am asking! :) Thank you, my math-english is so bad, I am really having troubles setting the words right.

Hint: Start by writing out some equations which combine more than one of the three basic equations (and call them p q and r rather than k1 k2 and k3 … it's easier to type ).
What do you get?

not much :) i can see that 2|a, 3|a, 5|a .. but i can't go passed that... i tried doing a more equations with the given statements, but i seem to be going in circle.
I know I need to use something with divisors or GDC, i just can't see the connection. The GDC(2,3,5)=1, which doesent give me much either.
I wish there was given at least one more statement or something. I am totally stuck :(

 

[tiny-tim]

not much :) i can see that 2|a, 3|a, 5|a .. but i can't go passed that

Hint: so call it 2^a3^b5^cd

 

[Eliva]

ohhh! i knew it had something to do with those numbers being simple. Ok, it's Euclids algorithm (which i said myself d'uh) ... so the number a can be factorized by simple numbers (2, 3 and 5 in this case) ... but how do I find their powers?
and just to make sure,

2^a3^b5^cd

in here 2^a this "a" here is different than the number a we are looking for, right??

so a= 2^x3^y5^zd
what is the last digit d there??, would it be 7? is there suppose to be one more multiplier, or i need to only use 2, 3 and 5?? andddd, how do i find the powers (lets call them x,y,z, for no confusion) ??

 

[Tedjn]

Given the way you have written a, what is a/2, a/3, a/5? What must they equal?

 

[Eliva]

Given the way you have written a, what is a/2, a/3, a/5? What must they equal?

In the natural numbers, N, find the lowest such that half of it is a perfect square, one-third is a perfect cube, and one fifth is a perfect fifth power.

thats the assignment

 

[tiny-tim]

No, Tedjn meant a= 2^x-13^y5^zd,

a/3 = … ?

a/5 = … ?

(and d can be any number, not divisible by 2 3 or 5)