Factor Craze 
Here’s a Problem to Ponder that will help your students ease back into the groove at the start of the school year. Students usually know that prime numbers have exactly two factors —1 and the number itself. But, which numbers have exactly three factors? Exactly four factors? Exactly five factors? Exactly six factors?

The general extension: Given a positive integer, n, how can we tell exactly how many factors it has?

Some Ponderings: 

The August Problem to Ponder posed the following question: Which numbers have exactly three factors? Exactly four factors? Exactly five factors? Exactly six factors?
Given a positive integer n, how can we tell exactly how many factors it has?

Students might start to look for examples of numbers that have exactly 3, 4, 5, or 6. factors. Factors travel in pairs. For example, primes have two factors, 1 and themselves. Consider a number like 24—its factors appear in pairs, 1 and 24, 2 and 12, 3 and 8, 4 and 6.

Exactly three factors 
Students often move quickly to looking at the prime factors of a number as they start this problem. Because factors travel in pairs, the only way in which we could have exactly 3 different factors is if we had a number with a repeated factor, like 3 x 3. In fact, 9 does have exactly three factors: 1, 3, and 9. And 4 also has exactly three factors: 1, 2, and 4. A first conjecture that students sometimes make is that every perfect square has three factors—but of course that’s not true. For example, 36 has many more than three factors. However, every number that is a square of a prime p will have exactly three factors: 1, p, and p².

Exactly four factors
Some examples of numbers with exactly four factors are 6, 15, and 21.  For 6, the factors are 1, 2, 3, and 6, and for 15, the factors are 1, 3, 5, and 15. Similarly, for 21, the factors are 1, 3, 7, and 21. Students soon see that numbers of the form p x q, where p and q are distinct primes (p ≠ q) will have only four factors: 1, p, q, and pq. However, these aren’t the only numbers that have exactly four factors. For example, 8 has 1, 2, 4, and 8 as factors, and 8 = 2³. Any number that is of the form p³ for some prime p also has exactly four factors: 1, p, p² and p³.

Exactly five factors 
One pattern that is emerging so far involves powers of primes. We know that any number that is of the form p⁴ for some prime p also will have exactly five factors: 1, p, p², p³, and p⁴. And in general, numbers of the form pⁿ will have n+1 factors.

Exactly six factors
In addition to numbers of the form p⁵ for some prime p, numbers like 20 = 2² x 5 will have six factors: 1, 2, 2², 5, 2 x 5, and 2² x 5. Thus, numbers that have exactly six factors are either a 5th prime power (like 32 = 2⁵) or of the form p² x q, where p and q are primes.

The total number of factors is related to the prime factorization of an integer. The general question is, If we know the prime factorization of a number, how can we tell exactly how many factors it has? We’ll return with more ponderings on Factor Craze in next month’s column!

Further Ponderings:

The August Problem to Ponder posed the following question: Which numbers have exactly three factors? Exactly four factors? Exactly five factors? Exactly six factors?

Given a positive integer n, how can we tell exactly how many factors it has?

In last month’s Ponderings, we found the following solutions for the first few cases.

Every number that is a square of a prime p has exactly three factors: 1, p, and p².

Numbers of the form p³ for some prime p, or of the form p x q, where p and q are distinct primes, have exactly four factors.

Numbers of the form p⁴ for some prime p have exactly five factors: 1, p, p², p³, and p⁴.

Numbers of the form p⁵ for some prime p, or of the form p² x q, where p and q are primes, have exactly six factors.

The total number of factors is related to the prime factorization of an integer, since all factors of a number are built from combinations of its prime factors. We already know that if an integer is a prime power—say, pⁿ—it has exactly (n + 1) factors: 1, p, p², p³, …, pⁿ. If an integer has two prime powers in its decomposition—say, pⁿ x q^m, p ≠ q—then all of its factors are composed of combinations of powers p and q. The collections of powers are respectively {1, p, p², p³, … pⁿ} and {1, q, q², q³, … q^m}. There are (n +1) choices of factors from the first of these sets, and (m + 1) choices of factors from the second set, and thus, there are (n + 1) x (m + 1) total possible factors for any integer that has a prime factorization of the form pⁿ x q^m. (You might consider 1 in each of the sets as p⁰ and q⁰, respectively).

For example, 72 = 2³ x 3². So, using the exponents, we know that there will be (3 + 1) x (2 + 1) = 12 different factors of 72—namely,

            1, 2, 2², 2³, 3, 3², (2 x 3), (2 x 3²), (2² x 3), (2² x 3²), (2³ x 3), and (2³ x 3²).

Extending this idea, if an integer is of the form pⁿ x q^m x r^t, where p, q, and r are distinct primes, then it has (n + 1) x (m + 1) x (r + 1) total factors, and so forth for prime decompositions with more prime powers in them.

We entered this “pondering” by considering a few special cases that are accessible to younger students, and these us led to extensions and to a general case that are challenging for older students. Such problems provide ponderings for students of many age levels.