A3. If you calculate $6 \times 5 \times 4 \times 3 \times 2 \times 1$, you get 720. How many divisors does the number 720 have? (A divisor of a number $n$ is a positive integer by which $n$ is divisible. Examples: the divisors of 6 are 1, 2, 3, and 6; the divisors of 11 are 1 and 11.)
A) 6
B) 8
C) 20
D) 30
E) 36
A3. D) 30 The number 720 contains only the prime factors 2, 3, and 5. The prime number 2 appears four times (once in 2, twice in 4, and once in 6), the prime number 3 appears twice (in 3 and in 6), and 5 appears once. The divisors without the factor 3 and 5 are 1, 2, 4, 8, and 16. The divisors with one factor 3 and no factor 5 are 3, 6, 12, 24, and 48. With the factor 9 and no factor 5: 9, 18, 36, 72, and 144. Without the factor 5, there are thus 15 divisors. Each of these 15 divisors, when multiplied by 5, gives the remaining 15 divisors. In total, there are 30 divisors.

Alternative solution: Each divisor of \(720 = 2^{4} \times 3^{2} \times 5^{1}\) is of the form \(2^{a} \times 3^{b} \times 5^{c}\) with five possibilities for the factor \(2^{a}\) (from \(2^{0}\) to \(2^{4}\)), three possibilities for \(3^{b}\) (\(3^{0}\), \(3^{1}\), and \(3^{2}\)), and two possibilities for \(5^{c}\) (\(5^{0}\) and \(5^{1}\)). Combined, this gives \(5 \times 3 \times 2 = 30\) possibilities.