Chapter
Divisor Counting and Special Numbers
Counting divisors with prime factorization, squarefree divisors, products of divisors, factorial exponents, and special number patterns.
Theory
1. Counting Divisors
If \(n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}\), then every divisor is obtained by choosing exponents \(0\le e_i\le a_i\). Thus \(d(n)=(a_1+1)\cdots(a_k+1)\).
2. Squarefree Divisors
A squarefree divisor has each prime exponent equal to \(0\) or \(1\).
3. Products and Factorials
Pair divisors as \(d\) and \(n/d\). For factorials, count prime powers with floors.
Examples
Example 1. Count Divisors of 3600
This is the basic divisor-counting model.
Problem.
How many positive divisors does \(3600\) have?
Solution.
Since \(3600=2^4\cdot3^2\cdot5^2\), the number of divisors is \(5\cdot3\cdot3=45\).
Example 2. Squarefree Divisors
This reformulates the squarefree-divisor idea.
Problem.
How many divisors of \(2^3\cdot3^2\cdot5\cdot7\) are not divisible by any square greater than \(1\)?
Solution.
There are four available primes, and each is either chosen or not chosen. The answer is \(2^4=16\).
Example 3. Seven Divisors Then Squared
This tests how divisor counts determine exponent patterns.
Problem.
A natural number \(n\) has exactly \(7\) positive divisors. How many positive divisors does \(n^2\) have?
Solution.
If \(n=p^6\), then \(n^2=p^{12}\), so \(d(n^2)=13\).
Example 4. Power of Five in a Factorial
This is the standard factorial-exponent calculation.
Problem.
Find the largest exponent \(e\) such that \(5^e\mid100!\).
Solution.
The exponent is \(\lfloor100/5\rfloor+\lfloor100/25\rfloor=20+4=24\).
Problems
Problems
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Ladders
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