Math Is Fun Forum

shocamefromebay

Member

Registered: 2007-05-30

Posts: 103

Number of Factors.

There is this problem I cannot figure out. There are two parts to it, but I have only figured out one of them.

The notation, "

," signifies the number of factors that N has. For example,

is equal to 3 because it has 3 factors, which are 1,2 and 4.

is equal to 6 because it has 6 factors, which are 1, 2, 3, 4, 6, and 12.

The first part, the one that I could figure out, is as follows:

If b is an even number and both a and b are integers greater than ten but less than 99, find an expression for x using the variables a and b to make the statement above true.

The answer that I found is as follows:

Here is the second part.

If b is an odd number and both a and b are integers greater than ten but less than 99, find an expression for x using the variables a and b to make the statements above true.

I can't quite figure out the last part. Can anyone please help me?

3lawrence

Member

Registered: 2009-06-02

Posts: 4

Re: Number of Factors.

Sorry.i dunno.

shocamefromebay

Member

Registered: 2007-05-30

Posts: 103

Re: Number of Factors.

Can no one help me with this? This problem has been bugging me for a few months now and I still can't figure it out.

Last edited by shocamefromebay (2009-06-07 10:15:08)

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Number of Factors.

I think the requirement in the second part should be that a is odd rather than b.

You have probably already shown in the first part that

In general

will depend on the prime factorisation of

.
If we assume that

is a prime power

then

so we should choose

and

can be any prime.
These are not the only solutions, but I do think they are the simplest.

shocamefromebay

Member

Registered: 2007-05-30

Posts: 103

Re: Number of Factors.

I don't mean to be rude, but I reread the question and it does say that b is odd.

Also, I may be wrong, and if I am please tell me, but I am not sure if that solution works all the time. What if (x+1) is not a prime power?

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Number of Factors.

I'm not questioning you ability to read or copy, but suggesting that whoever set the problem may have made a mistake.
If

then it turns out that

is odd, so there can be no solution when a is even.

The problem is to find an expression for x in terms of a and b such that a certain relationship between these three quantities holds. The value of x is not uniquely determined by a and b, and in my solution I only try to find some value of x that works. It is always possible to choose x so that x+1 is a prime power, and I choose to do so.

As an example, suppose a=3 and b=1. For any prime p,

and

. Hence we can choose x to be p-1 for and prime p.

Last edited by Avon (2009-06-07 19:37:14)

shocamefromebay

Member

Registered: 2007-05-30

Posts: 103

Re: Number of Factors.

If

then it turns out that

is odd, so there can be no solution when a is even.

If

,

is not always odd.

For example, let x = 4.

I think that the question is looking for a general solution that works all the time. The solution you provided does work, but only when x+1 is a prime power. There still needs to be a solution for when x+1 is not a prime power.

I think that there may need to be many separate solutions to find the all the possible solutions for this problem, like an expression for x when a is odd and a different one when a is even. The question states that b needs to be odd and a can be anything as long as both a and b are integers greater than ten and less than 99.

noobard

Member

Registered: 2009-06-07

Posts: 28

Re: Number of Factors.

Hmm I got a solution for this as follows :

We know that a number can be expressed a s a product of its factors.............
10^x can be written as 2^x * 5^x...................and we know the number of factors comes out to be
(x+1)^2 ...................so (x+1)^2=a^b
we can solve this quadratic equation and since b is even we can take the positive square root and hence get the answer

---

Everything that has a begining has an EnD!!!

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Number of Factors.

If

,

is not always odd.

For example, let x = 4.

which is odd.
It is not too hard to prove that any non-zero square has an odd number of factors.

If b is an odd number and both a and b are integers greater than ten but less than 99, find an expression for x using the variables a and b to make the statements above true.

Suppose I write

This is an expression for x using the variables a and b.

So this expression for x makes the statements true.

I really can't see how I have failed to find an expression for x using the variables a and b to make the given statements true.
The only problem I can see with this solution is that x is not an integer when a is even, but no solution is possible in that case.

I think that there may need to be many separate solutions to find the all the possible solutions for this problem

As I admitted in my first reply, the solutions I have given are not the only solutions, so I completely agree with you that there is more work to be done in order to find all possible solutions. However, until we can agree that I have provided some solutions to this problem I have no interest in looking for others.

noobard

Member

Registered: 2009-06-07

Posts: 28

Re: Number of Factors.

oops i wrote it for the first part....zzzzzzzzzz

ll try ma way for the second part............

---

Everything that has a begining has an EnD!!!

alyssakstidham

Member

Registered: 2009-05-22

Posts: 1

Re: Number of Factors.

That Worked