Problem:
When a standard 6-sided die is rolled n times, the product of the n numbers rolled can be any of 936 possible values. What is n?
Answer Choices:
A. 11
B. 6
C. 8
D. 10
E. 9
Solution:
The product of the results of the rolls takes the form 2a3b5c, and every roll contributes at most two prime factors to the product. No roll contributes more than one factor of 5 , so 0≤c≤n.
No roll contributes more than one factor of 3 , and no roll can contribute both a 3 and a 5 , so 0≤b≤ n−c. No roll contributes more than two factors of 2 , no roll that contributes a 3 can contribute more than one factor of 2 , and no roll can contribute both a 2 and a 5 , so 0≤a≤2n−b−2c. Each ordered triple (a,b,c) determines a different product, so the number of possible products is
c=0∑n​b=0∑n−c​(2n−b−2c+1)​=c=0∑n​(b=0∑n−c​(2n−2c+1)−b=0∑n−c​b)=c=0∑n​((2(n−c)+1)(n−c+1)−21​(n−c)(n−c+1))=c=0∑n​((2c+1)(c+1)−21​c(c+1))=c=0∑n​(23​c2+25​c+1)=23​⋅6n(n+1)(2n+1)​+25​⋅2n(n+1)​+(n+1)=2(n+1)2(n+2)​​
Note that 936=2122⋅13​, so n=(A)11​.
The problems on this page are the property of the MAA's American Mathematics Competitions