If p,r, and s are three different prime numbers greater than 2 and n = p*r*s, how many positive factors, including 1 and n, does n have?

8

here are two ways to look at it,

1. how many subsets can you form with 3 elements, including the null set (the factor 1), and the set of all elements (the n) -----> 2^3 = 8

or

2. we can use the p in 2 ways, that is, we can either include it or we can not include it, so 2 x ...
we can use the r in 2 ways, that is, we can either include it or we can not include it, so 2 x 2 x ..
we can use the s in 2 ways, that is, we can either include it or we can not include it, so 2 x 2 x 2

2 x 2 x 2 = 8,
this would include not taking any, which might correspond to the factor 1, and includes taking all 3 of them, which corresponds to n.

try it with the 3 numbers 3,5 and 7

factors would be
1
2
5
7
3*5
3*7
5*7 and finally
3*5*7, count them, there are 8

To find the number of positive factors of a given number, we first need to find the prime factorization of that number. In this case, we know that n is the product of three different prime numbers: p, r, and s.

As all three numbers are prime, their prime factorizations will only include themselves. So, the prime factorization of n will be: n = p * r * s.

Now, let's consider the factors of n. Each factor of n can be expressed in the form of p^a * r^b * s^c, where a, b, and c are non-negative integers.

Since we have three prime factors, each factor can be chosen with or without any of the primes. This means that we have (a+1) choices for the power of p, (b+1) choices for the power of r, and (c+1) choices for the power of s.

Therefore, the total number of factors of n, including 1 and n itself, is equal to (a+1) * (b+1) * (c+1).

In this case, since p, r, and s are distinct prime numbers, all the powers (a, b, c) will be equal to 1. So, the number of positive factors of n is (1+1) * (1+1) * (1+1), which simplifies to 2*2*2 = 8.

Therefore, the number n has 8 positive factors, including 1 and n.