To find the factors of 4000, I need to factorize it into prime factors. I know that 4000 is divisible by 2, so I can start by dividing it by 2 repeatedly until I get an odd number. I get 4000 = 2^5 * 125. Now, 125 is divisible by 5, so I can divide it by 5 repeatedly until I get 1. I get 125 = 5^3. So, 4000 = 2^5 * 5^3. This means that any factor of 4000 must be of the form 2^a * 5^b, where a and b are non-negative integers and a <= 5 and b <= 3. To be a multiple of 100, a factor must have at least two factors of 2 and two factors of 5. So, a must be at least 2 and b must be at least 2. For a, I have four choices: 2, 3, 4, or 5. For b, I have two choices: 2 or 3. Therefore, the number of factors of 4000 that are multiples of 100 is 4 * 2 = 8. 
How many positive factors of 4000 are multiples of 100?