1. Consider the number 539. Note that 539 = 72 x 11.(a) How many different values of a can you choose so that 7ª is a factor of 539? Explainhow you arrived at your answer.(b) How many different values of b can you choose so that 11 is a factor of 539? Explainhow you arrived at your answer.(c) As discussed in class, we know that any factor of 539 must be made up of sevens andelevens. In other words, any factor of 539 can be expressed as 7" x 11º for some r andy. Use this fact and your work in the previous two parts to determine how many factors539 will have without actually finding any of them. Explain how you arrived at youranswer.

A number is considered a factor of another number only when the first number divides the second number perfectly. The prime factors of a number are the prime numbers that divide the number perfectly.(a) From the given expression 539=72×11, it is observed that the maximum number of 7s a factor of 539 can have is 2. So, the different values of a that can be chosen are 0, 1, and 2. (b) From the given expression 539=72×11, it is observed that the maximum number of 11s a factor of 539 can have is 1. So, the different values of b that can be chosen are 0 and 1.(c) From the obtained information, the possible number of values of x is 3, and the possible number of values of y is 2. So, the total number of possible factors n of 539 are as follows:- n=3×2=6 Thus, there are 6 factors possible for 539.