Difference between revisions of "Divisor function"

Revision as of 10:36, 2 August 2006

The divisor function is denoted $\sigma_k(n)$ and is defined as the sum of the $k$ th powers of the divisors of $n$ . Thus $\sigma_k(n) = \sum_{d|n}d^k = d_1^k + d_2^k + \cdots + d_r^k$ where the $d_i$ are the positive divisors of $n$ .

Counting divisors

Note that $\sigma_0(n) = d_1^0 + d_2^0 + \ldots + d_r^0 = 1 + 1 + \ldots + 1 = r$ , the number of divisors of $n$ . Thus $\sigma_0(n) = d(n)$ is simply the number of divisors of $n$ .

Example Problems

Demonstration

Consider the task of counting the divisors of 72.

First, we find the prime factorization of 72: $\displaystyle72=2^{3} \cdot 3^{2}.$

Since each divisor of 72 can have a power of 2, and since this power can be 0, 1, 2, or 3, we have 4 possibilities. Likewise, since each divisor can have a power of 3, and since this power can be 0, 1, or 2, we have 3 possibilities. By an elementary counting principle, we have $3\cdot 4=12$ divisors.

We can now generalize. Let the prime factorization of $n$ be $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ . Any divisor of $n$ must be of the form $p_1^{f_1}p_2^{f_2} \cdots p_k^{e_k}$ where the $\displaystyle f_i$ are integers such that $0\le f_i \le e_i$ for $i = 1,2,\ldots, k$ . Thus, the number of divisors of $n$ is $\sigma_0(n) = (e_1+1)(e_2+1)\cdots (e_k+1)$ .

Introductory Problems

2005 AMC 10A Problem 15

Sum of divisors

The sum of the divisors, or $\sigma_1(n)$ , is given by

$\sigma_1(n) = (1 + p_1 + p_1^2 +\cdots p_1^{e_1})(1 + p_2 + p_2^2 + \cdots + p_2^{e_2}) \cdots (1 + p_n + p_n^2 + \cdots + p_n^{e_n}).$

@@ Line 5: / Line 5: @@
 Note that <math>\sigma_0(n) = d_1^0 + d_2^0 + \ldots + d_r^0 = 1 + 1 + \ldots + 1 = r</math>, the number of divisors of <math>n</math>.  Thus <math>\sigma_0(n) = d(n)</math> is simply the number of divisors of <math>n</math>.
-=== Example ===
+=== Example Problems  ===
+==== Demonstration ====
 Consider the task of counting the divisors of 72.
@@ Line 13: / Line 14: @@
 We can now generalize.  Let the prime factorization of <math>n</math> be <math>p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}</math>.  Any divisor of <math>n</math> must be of the form <math>p_1^{f_1}p_2^{f_2} \cdots p_k^{e_k}</math> where the <math>\displaystyle f_i </math> are integers such that <math>0\le f_i \le e_i</math> for <math>i = 1,2,\ldots, k</math>.  Thus, the number of divisors of <math>n</math> is <math>\sigma_0(n) = (e_1+1)(e_2+1)\cdots (e_k+1)</math>.
+==== Introductory Problems ====
+* [[2005_AMC_10A_Problems/Problem_15 | 2005 AMC 10A Problem 15]]
 == Sum of divisors ==
 The sum of the divisors, or <math>\sigma_1(n)</math>, is given by <center><math> \sigma_1(n) = (1 + p_1 + p_1^2 +\cdots p_1^{e_1})(1 + p_2 + p_2^2 + \cdots + p_2^{e_2}) \cdots (1 + p_n + p_n^2 + \cdots + p_n^{e_n}).</math> </center>
 == See also ==
 * [[Divisibility]]
 * [[Number theory]]

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