# Need desparate help on this question concerning finding a positive integer

1. Mar 14, 2005

### clueles

fi have no idea what to do and i tried posting it on another forum and nobody replied so please help me! thank you so much!

find a positive integer n so that 40n is a fifth power (of an integer) 500n is a sixth power, and 200n is a seventh power, or explain why it is impossible to do so (hint: let x = e2(n) and y = e5(n) use the given assumptions about n to find systems of linear congruences that x and y must satisfy. you may leave n in terms of its canonical prime factorization. show as much work as possible

2. Mar 14, 2005

### Muzza

40 = 2^3 * 5, 500 = 2^2 * 5^3, 200 = 2^3 * 5^2.

For simplicity, assume n = 2^x * 5^y for some natural numbers x, y.

Then 40n = 2^(x + 3) * 5^(y + 1). If this is to be a fifth power, both exponents must be divisible by 5, i.e.

x + 3 == 0 (mod 5),
y + 1 == 0 (mod 5).

You can find similar requirements for 500n and 200n. Then you solve this system of equations.

Last edited: Mar 14, 2005
3. Mar 14, 2005

### clueles

Thank you for answering my question. I just need some clarification though. When you have the
x + 3 == 0 (mod 6),
y + 1 == 0 (mod 6).

do you solve the congruence x + 3 == 0(mod6) and then you solve the second y+1 == 0(mod6) and once you do you subsitute the remainder into the n = 2^x * 5^y

For the 500n = 2^(x + 2) * 5^(y + 3).
x+2 == 0(mod6)
y+3 ==0(mod6)

what happens when you solve all the equations how do you know what n would be if you have different numbers?

4. Mar 14, 2005

### Muzza

I made a mistake in my original post, 40n was supposed to be a fifth power. The equations need to be changed accordingly.

No. You must find x, y such that all six equations are satisfied /simultaneously/:

x + 3 == 0 (mod 5),
y + 1 == 0 (mod 5),
x + 2 == 0 (mod 6),
y + 3 == 0 (mod 6),
x + 3 == 0 (mod 7),
y + 2 == 0 (mod 7).

Have you heard about the Chinese remainder theorem? (This system can be solved without explicitely using the CRT though).

After you have found a solution, you plug it in to n = 2^x * 5^y. That's how you find out what n "is" (but n is not uniquely determined, i.e. there are infinitely many n that will work. The original problem was to find /an/ n that worked, not all n).

5. Mar 14, 2005

### clueles

Question again

how can you use the chinese remainder theorem if you have an x and y and you want to do it simulatenously?

6. Mar 14, 2005

### Muzza

Um, by first solving

x + 3 == 0 (mod 5)
x + 2 == 0 (mod 6)
x + 3 == 0 (mod 7)

using the CRT, and then doing the same to

y + 1 == 0 (mod 5)
y + 3 == 0 (mod 6)
y + 2 == 0 (mod 7).

7. Mar 14, 2005

### clueles

to use the chinese remainder theorem wouldn't you have to rewrite the equation s for the x's as
x==-3(mod5)
x==-2(mod6)
x==-3(mod7)

so then we can rewrite it using the q's. and once we figure out what the general form of it the x would be the remainder of it