Guide 1: Chapter 4: Question: 11 (pg. 58, 4th edition)

[lionelpq]

The answer to question 11 says "As we learned in Chapter 1, there are (4+1)(1+1)=10 different factors of 2^4*3." What is this formula refered to in Chapter 1 that gets us to 10? I could not find it.

Thanks!

[phoenix.rkchr]

The answer to question 11 says "As we learned in Chapter 1, there are (4+1)(1+1)=10 different factors of 2^4*3." What is this formula refered to in Chapter 1 that gets us to 10? I could not find it.

Thanks!

The formula to find the total number of factors of any number N is:

(a+1)(b+1)(c+1)........

where a, b, c, ..... are the powers of prime factors of the number N.

e.g. N = 72
prime factors of 72= 2X2X2X3X3 = 2^3 X 3^2

here a = 3 and b = 2

So number of factors of 72 = (3+1)(2+1) = 12

[lionelpq]

Apreciate it. Thanks

[tim]

Thanks, Phoenix..