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Revision as of 01:23, 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

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