Factor Craze
Here’s a Problem to Ponder thatwill help your students ease back into the groove at the start of theschool year. Students usually know that prime numbers have exactly twofactors —1 and the number itself. But, which numbers have exactly threefactors? Exactly four factors? Exactly five factors? Exactly sixfactors?
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 followingquestion: Which numbers have exactly three factors? Exactly fourfactors? 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 haveexactly 3, 4, 5, or 6. factors. Factors travel in pairs. For example,primes have two factors, 1 and themselves. Consider a number like 24—itsfactors appear in pairs, 1 and 24, 2 and 12, 3 and 8, 4 and 6.
Exactly three factors
Students often movequickly to looking at the prime factors of a number as they start thisproblem. Because factors travel in pairs, the only way in which we couldhave exactly 3 different factors is if we had a number with a repeatedfactor, 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 firstconjecture that students sometimes make is that every perfect square hasthree factors—but of course that’s not true. For example, 36 has manymore than three factors. However, every number that is a square of aprime p will have exactly three factors: 1, p, and p2.
Exactly four factors
Some examples of numberswith 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, for21, the factors are 1, 3, 7, and 21. Students soon see that numbers ofthe 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 ofan integer. The general question is, If we know the prime factorizationof a number, how can we tell exactly how many factors it has? We’llreturn 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? Exactlyfive 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 ofan integer, since all factors of a number are built from combinations ofits prime factors. We already know that if an integer is a primepower—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 qm, 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, … qm}. 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 qm. (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 qm x rt, 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 thatare accessible to younger students, and these us led to extensions andto a general case that are challenging for older students. Such problemsprovide ponderings for students of many age levels.
