How many five-digit numbers exist such that the product of their digits is 400?	Answers:
A) 60  B) 120  C) 180  D) 30  E) 75
We're asked to name all of the 5-digit numbers whose product-of-digits is 400. We can use prime-factorization to figure out the digits, BUT this comes with a "twist".....NOT ALL of the digits have to be primes....   400 = (25)(16)   The 25 = (5)(5) and there's no way around that. Every option MUST have two 5s as two of the digits.   The 16 can be broken down in a few different ways though:   16 = (1)(2)(8)   16 = (1)(4)(4)   16 = (2)(2)(4)   Each of those options has to be accounted for.   So, we have 3 sets of 5 digits:   1) 1, 2, 8, 5, 5 --&gt; This gives us 5!/2! arrangements = 120/2= 
<gadget id="calculator">120/2</gadget>
<output>60</output>
60 different 5-digit numbers.   2) 1, 4, 4, 5, 5 --&gt; This gives us 5!/(2!2!) arrangements= 120/4= 
<gadget id="calculator">120/4</gadget>
<output>30</output>
30 different 5-digit numbers.   3) 2, 2, 4, 5, 5 --&gt; This gives us 5!/(2!2!) arrangements= 120/4= 
<gadget id="calculator">120/4</gadget>
<output>30</output>
30 different 5-digit numbers.   60 + 30 + 30= 
<gadget id="calculator">60 + 30 + 30</gadget>
<output>120</output>
120 different 5-digit numbers   Final Answer:120
<result>B</result>
How many five-digit numbers exist such that the product of their digits is 400?