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2020 AMC 12B Problems/Problem 24

Revision as of 17:29, 9 February 2020 by Rzlng (talk | contribs) (→‎Solution 1)

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 1

To make a product of $n$, we can either have just $n$, or we can have a divisor $d$ of $n$, $1 < d < n$, followed by a way to make a product of $\frac{n}{d}$. Thus, $D(n) = 1 + \sum_{d | n} D \left(\frac{n}{d} \right)$ for all possible $d$. It's easy to calculate $D(n)$ for all powers of $2$, since powers of $2$ only have powers of $2$ as divisors. We have \[D(2) = 1\] \[D(4) = 2\] \[D(8) = 4\] \[D(16) = 8\] \[D(32) = 16\] Now we will calculate $D(n)$ for all $n$ in the form $3 \cdot 2^k$, for $k \geq 0$. Note that each divisor of $3 \cdot 2^k$ is either of the form $2^j$ or $3 \cdot 2^j$. Therefore, to calculate each $D(3 \cdot 2^k)$, we will sum the first $k$ values in both the tables we created, and add $1$. We have \[D(3) = 1\] \[D(6) = 3\] \[D(12) = 8\] \[D(24) = 20\] \[D(48) = 48\] \[D(96) = \boxed{\textbf{(A) }112}\]

Solution 2

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 $\boxed{\textbf{(A) }112}$.

~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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