2009年全美数学竞赛（AMC）竞赛
有多少个三位数，其各个数位之积等于$$24$$？．
课内体系->知识点->数->有理数->有理数基础运算->有理数乘法->有理数乘法运算 美国AMC8->Knowledge Point->Number Theory->Factors and Multiples->Factors and Multiples Basics 美国AMC8->Knowledge Point->Number Theory->Place Value and Number Bases->Numbers and Digits
有多少个三位数，其各个数位之积等于$$24$$？  从最小到最大列出的数字，三位整数位$$138$$，$$146$$，$$226$$，$$234$$，$$226$$可按$$\frac{3！}{2！}=3$$方式排列，其他三种可按$$3！=6$$方式排列，有$$3+6(3)=\text{D}21$$个三位数的正整数．  With the digits listed from least to greatest, the $$3$$-digit integers are $$138$$, $$146$$, $$226$$, $$234$$, $$226$$ can be arranged in $$\frac{3!}{2!}=3$$ ways, and the other three can be arranged in $$3!= 6 $$ways. There are $$3+6(3)=21$$,$$3$$-digit positive integers. 