1. 题目

12000的因数有多少个？

2. 枚举法求解

2.1 分解质因数

$$\begin{gathered} 2\underline{|12000} \\ 2\underline{|6000} \\ 2\underline{|3000} \\ 2\underline{|1500} \\ 2\underline{|750} \\ 3\underline{|375} \\ 5\underline{|125} \\ 5\underline{|25} \\ 5 \end{gathered}$$
整理得：
$12000=2^5\times 3^1\times 5^3$

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2.2 枚举

序号	质因数
1	$1=2^0\times 3^0\times 5^0$
2	$5=2^0\times 3^0\times 5^1$
3	$25=2^0\times 3^0\times 5^2$
4	$125=2^0\times 3^0\times 5^3$
5	$3=2^0\times 3^1\times 5^0$
6	$15=2^0\times 3^1\times 5^1$
7	$75=2^0\times 3^1\times 5^2$
8	$375=2^0\times 3^1\times 5^3$
9	$2=2^1\times 3^0\times 5^0$
10	$10=2^1\times 3^0\times 5^1$
11	$50=2^1\times 3^0\times 5^2$
12	$250=2^1\times 3^0\times 5^3$
13	$6=2^1\times 3^1\times 5^0$
14	$30=2^1\times 3^1\times 5^1$
15	$150=2^1\times 3^1\times 5^2$
16	$750=2^1\times 3^1\times 5^3$
17	$4=2^2\times 3^0\times 5^0$
18	$20=2^2\times 3^0\times 5^1$
19	$100=2^2\times 3^0\times 5^2$
20	$500=2^2\times 3^0\times 5^3$
21	$12=2^2\times 3^1\times 5^0$
22	$60=2^2\times 3^1\times 5^1$
23	$300=2^2\times 3^1\times 5^2$
24	$1500=2^2\times 3^1\times 5^3$
25	$8=2^3\times 3^0\times 5^0$
26	$40=2^3\times 3^0\times 5^1$
27	$200=2^3\times 3^0\times 5^2$
28	$1000=2^3\times 3^0\times 5^3$
29	$24=2^3\times 3^1\times 5^0$
30	$120=2^3\times 3^1\times 5^1$
31	$600=2^3\times 3^1\times 5^2$
32	$3000=2^3\times 3^1\times 5^3$
33	$16=2^4\times 3^0\times 5^0$
34	$80=2^4\times 3^0\times 5^1$
35	$400=2^4\times 3^0\times 5^2$
36	$2000=2^4\times 3^0\times 5^3$
37	$48=2^4\times 3^1\times 5^0$
38	$240=2^4\times 3^1\times 5^1$
39	$1200=2^4\times 3^1\times 5^2$
40	$6000=2^4\times 3^1\times 5^3$
41	$32=2^5\times 3^0\times 5^0$
42	$160=2^5\times 3^0\times 5^1$
43	$800=2^5\times 3^0\times 5^2$
44	$4000=2^5\times 3^0\times 5^3$
45	$96=2^5\times 3^1\times 5^0$
46	$480=2^5\times 3^1\times 5^1$
47	$2400=2^5\times 3^1\times 5^2$
48	$12000=2^5\times 3^1\times 5^3$

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2.3 整理分析

与$2$相关的数有$2^0$、$2^1$、$2^2$、$2^3$
、$2^4$、$2^5$ 共6个数
与$3$相关的数有$3^0$、$3^1$ 共2个数
与$5$相关的数有$5^0$、$5^1$ 、$5^2$ 、$5^3$ 共4个数。
将他们分类组合有6×2×4种可能，所以我们得出以下公式：

$$C=\prod _{i=1}^n(Pn+1)$$
其中：
$C$ 为所有因数个数；
$P_n$ 为分解质因数后质因数的指数；
$n$ 为不同质因数的个数；
$\prod$ 为连乘或连积符号。

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3. 公式求解

因数公式
$$C=\prod _{i=1}^n(Pn+1)$$

分解因数：
$12000=2^5\times 3^1\times 5^3$
有3个不同的质因数，所以：$n=3$
3个不同的质因数的指数分别是：
$$\{\begin{array}{ll}{P}_1=5& 第1个质因数2的指数\\ P_2=1& 第2个质因数3的指数\\ P_3=3& 第3个质因数5的指数\end{array}$$

$$\begin{array}{c}\begin{array}{rl}\therefore C& =\prod _{i=1}^n(Pn+1)\\ & =(P_1+1)\times (P_2+1)\times (P_3+1)\\ & =(5+1)\times (1+1)\times (3+1)\\ & =6\times 2\times 4\\ & =48\end{array}\end{array}$$

4. 扩展

4.1 因式分解

分解质因数

/// <summary>
/// 分解质因数函数
/// </summary>
/// <param name="number"></param>
/// <returns></returns>
static string DecomposingPrimeFactors(int number)
{int N = number;List<int> temps = new List<int>();int primeNumber = 2;DecomposingPrimeFactors(ref N, ref primeNumber, ref temps);return FormatPowerlaw(number,temps);
}
/// <summary>
/// 递归查找质因数
/// </summary>
/// <param name="N">传入需要分解的数</param>
/// <param name="primeNumber">传入第一个质数1</param>
/// <param name="temp">存储找到的质因数</param>
static void DecomposingPrimeFactors(ref int N, ref int primeNumber, ref List<int> temp)
{if (N <= 1){return;}if (N % primeNumber == 0){temp.Add(primeNumber);N = N / primeNumber;DecomposingPrimeFactors(ref N, ref primeNumber, ref temp);}else{primeNumber++;DecomposingPrimeFactors(ref N, ref primeNumber, ref temp);}
}

格式化输出结果

/// <summary>
/// 格式化质因数-乘式
/// N=a×a×a×a...×b×b×b×b×b...×c×c×c×c×c.....
/// </summary>
/// <param name="number"></param>
/// <param name="primeFactors"></param>
/// <returns></returns>
static string FormatMultiplicativeFormula(int number, List<int> primeFactors)
{return number + " = " + string.Join("×", primeFactors);
}
/// <summary>
/// 格式化质因数-指数式  
/// N=a^x1×b^x2×c^x3×.....
/// </summary>
/// <param name="number"></param>
/// <param name="primeFactors"></param>
/// <returns></returns>
static string FormatPowerlaw(int number, List<int> primeFactors)
{List<string> temps2 = new List<string>();foreach (var item in primeFactors.GroupBy(o => o)){int a = item.Key;int n = item.Count();temps2.Add($"{a}^{n}");}return number + " = " + string.Join("×", temps2);
}

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Console.WriteLine($"{DecomposingPrimeFactors(12000)}");

12000 = 2^5×3^1×5^3

4.2 因数个数

枚举法

/// <summary>
/// 枚举法计算因数个数
/// </summary>
/// <param name="N"></param>
/// <returns></returns>
static double CalculateFactorCount(double N)
{int cnt=0;for(double i=1; i<=N;i+=1){var v = Math.Pow(5, i);if(N % i == 0){cnt++;}}return cnt;
}

公式法

/// <summary>
/// 使用因数公式计算个数
/// </summary>
/// <param name="N"></param>
/// <returns></returns>
static double CalculateFactorCount2(int number)
{int cnt = 1;int N = number;List<int> primeFactors = new List<int>();int primeNumber = 2;DecomposingPrimeFactors(ref N, ref primeNumber, ref primeFactors);foreach (var item in primeFactors.GroupBy(o => o)){int n = item.Count();cnt *=(n+1);}return cnt;
}
/// <summary>
/// 递归查找质因数
/// </summary>
/// <param name="N">传入需要分解的数</param>
/// <param name="primeNumber">传入第一个质数1</param>
/// <param name="temp">存储找到的质因数</param>
static void DecomposingPrimeFactors(ref int N, ref int primeNumber, ref List<int> temp)
{if (N <= 1){return;}if (N % primeNumber == 0){temp.Add(primeNumber);N = N / primeNumber;DecomposingPrimeFactors(ref N, ref primeNumber, ref temp);}else{primeNumber++;DecomposingPrimeFactors(ref N, ref primeNumber, ref temp);}
}

调用

    Console.WriteLine($"{DecomposingPrimeFactors(12000)}");var Count = CalculateFactorCount(12000);Console.WriteLine($"枚举法：12000的因数有{Count}个");var Count2 = CalculateFactorCount2(12000);Console.WriteLine($"公式法：12000的因数有{Count2}个");Console.ReadLine();