Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 12B Problems/Problem 24"

(→‎Solution)
(→‎Solution)
Line 22: Line 22:
  
 
~FANYUCHEN20020715
 
~FANYUCHEN20020715
 +
 +
==See Also==
 +
  
 
{{AMC12 box|year=2020|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2020|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:23, 8 February 2020

Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product\[n = f_1\cdot f_2\cdots f_k,\]where $k\ge1$, the $f_i$ are integers strictly greater than $1$, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$, $2\cdot 3$, and $3\cdot2$, so $D(6) = 3$. What is $D(96)$?

$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$

Solution

Bash. Since $96=2^5\times 3$, for the number of $f_n$, we have the following cases:

Case 1: $n=1$, we have $\{f_1\}=\{96\}$, only 1 case.

Case 2: $n=2$, we have $\{f_1,f_2\}=\{3,2^5\}, \{6,2^4\},...,\{48,2\}$, totally $5\cdot 2!=10$ cases.

Case 3: $n=3$, we have $\{f_1,f_2,f_3\}=\{3,2^3,2^2\},\{3,2^1,2^4\},\{6,2^2,2^2\},\{6,2^3,2^1\}, \{12,2^2,2^1\},\{24,2,2\}$, totally $\frac{3!}{2!}\cdot 2+4\cdot 3!=30$ cases.

Case 4: $n=4$, we have $\{f_1,f_2,f_3,f_4\}=\{3,2^2,2^2,2\},\{3,2^3,2,2\},\{6,2^2,2,2\},\{12,2,2,2\}$, totally $\frac{4!}{2!}\cdot 3+\frac{4!}{3!}=40$ cases.

Case 5: $n=5$, we have $\{f_1,f_2,f_3,f_4,f_5\}=\{3,2^2,2,2,2\},\{6,2,2,2,2\}$, totally $\frac{5!}{3!}+\frac{5!}{4!}=25$ cases.

Case 6: $n=6$, we have $\{f_1,f_2,f_3,f_4,f_5,f_6\}=\{3,2,2,2,2,2\}$, totally $\frac{6!}{5!}=6$ cases.

Thus, add all of them together, we have $1+10+30+40+25+6=112$ cases. Put $\boxed{A}$.

~FANYUCHEN20020715

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_24&oldid=117361"