# Homework Help: Find smallest value of k in this equation

1. Jun 12, 2010

### thereddevils

1. The problem statement, all variables and given/known data

For these equations m,n,p,k are positive whole numbers greater than 1 .

n^(5/3)=m^(7/2)

nm=p^k

What is the smallest value that k can be?

(A) 6
(B) 11
(C) 31
(D) 41

2. Relevant equations

3. The attempt at a solution

some hints ?

2. Jun 12, 2010

### ocohen

Re: indices

try to get n in terms of m. Then see what k has to be such that p^k = nm

3. Jun 12, 2010

### thereddevils

Re: indices

that leaves me with p^k=m^(31/10)

and i am not told which is greater , p or m

4. Jun 12, 2010

### ocohen

Re: indices

you now get m = p^(10k/31) which means p^(10k/31) must be a whole number > 1.
So what does that mean for (10k/31)?

5. Jun 12, 2010

### thereddevils

Re: indices

erm 10k/31>0 ?

6. Jun 12, 2010

### ocohen

Re: indices

I would suspect that 10k/31 must be an integer so that p^(10k/31) is also an integer. I don't have any proof for this, but someone else might. As such it means that 10k must be a multiple of 31. Does that make sense?

EDIT: sorry this is incorrect.

7. Jun 12, 2010

### ocohen

Re: indices

If we are simply trying to satisfy the equation for the smallest k,
just solve it like this
m^(31/10) = p^k => m^(31/10k) = p
So if k = 6, we just need some m such that its 60th root is a whole number.
So let m = 2^60 or anything like that.
we have that n = m^(21/10) so n is also a whole number.

Does that make sense?

8. Jun 12, 2010

### thereddevils

Re: indices

well i have another thought on it ,

m=p^(10k/31)

For m is a positive whole number, 10k/31 must be an positive integer and p is known to be a positive whole number.

Therefore ,k must be a multiple of 31 where the smallest is 31 itself. SO the answer is obviously C

9. Jun 12, 2010

### ocohen

Re: indices

10k/31 does not need to be a positive integer.
Consider 9^(1/2)

10. Jun 13, 2010

### thereddevils

Re: indices

k is known to be a positive whole number from the question and none of the values of k for
1<k<31 is able to make the indices either a whole number or square roots , cube roots ,...until 31.