This page is for Math 355 students at Western Illinois University.
For example, 44,065 factors into (2^5)(3^4)(17^1).
44,065 = (2^5)(3^4)(17^1).
In general, a number, n, factors into k primes (with k exponents on the primes)
as (p1^e1)(p2^e2)(p3^e3)...(pk^ek). [This looks better using subscripts,
see example 4.41 on page 208.]
n = (p1^e1)(p2^e2)(p3^e3)...(pk^ek).
We want to count the number of factors.
First for 44,065: Factors will be of the form(2^_)(3^_)(17^_). A couple of example factors are (2^3)(17^1) and (2^5)(3^1)(17^1). In the blanks (exponents) we have to fill in numbers. Q: What are the possibilities for the three blanks? A: We can put in the blanks 0 to 5, 0 to 4, and 0 to 1, respectively (namely there are 6 choices, 5 choices, and 2 choices respectively). (Note that 2^0 would be 1, and that's OK.) Therefore, there are 6*5*2 different factors of 44,065. There are 60 factors of 44,065.
Generalizing, there are (e1 + 1)(e2 + 1)(e3 + 1)...(ek + 1) factors of n.
You can now get started on the assignment for section 4.5 but may not be able to do all of it until after we talk about it some more on 10/21/2002.
The assignment for section 4.5 is p. 210 #1,2,3,5,7,8.
James R. Olsen, Western Illinois University
E-mail: jr-olsen@wiu.edu
Page last updated: October 17, 2002