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| − | Let <math>D(n)</math> denote the number of ways of writing the positive integer <math>n</math> as a product<cmath>n = f_1\cdot f_2\cdots f_k,</cmath>where <math>k\ge1</math>, the <math>f_i</math> are integers strictly greater than <math>1</math>, 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 <math>6</math> can be written as <math>6</math>, <math>2\cdot 3</math>, and <math>3\cdot2</math>, so <math>D(6) = 3</math>. What is <math>D(96)</math>?
| + | #REDIRECT [[2020 AMC 10B Problems/Problem 25]] |
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| − | <math>\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184</math>
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| − | ==Solution 1==
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| − | To make a product of <math>n</math>, we can either have just <math>n</math>, or we can have a divisor <math>d</math> of <math>n</math>, <math>1 < d < n</math>, followed by a way to make a product of <math>\frac{n}{d}</math>. Thus, <math>D(n) = 1 + \sum D(\frac{n}{d})</math> for all possible <math>d</math>. It's easy to calculate <math>D(n)</math> for all powers of <math>2</math>, since powers of <math>2</math> only have powers of <math>2</math> as divisors. We have
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| − | <cmath>D(2) = 1</cmath>
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| − | <cmath>D(4) = 2</cmath>
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| − | <cmath>D(8) = 4</cmath>
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| − | <cmath>D(16) = 8</cmath>
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| − | <cmath>D(32) = 16</cmath>
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| − | Now we will calculate <math>D(n)</math> for all <math>n</math> in the form <math>3 \cdot 2^k</math>, for <math>k \geq 0</math>. Note that each divisor of <math>3 \cdot 2^k</math> is either of the form <math>2^j</math> or <math>3 \cdot 2^j</math>. Therefore, to calculate each <math>D(3 \cdot 2^k)</math>, we will sum the first <math>k</math> values in both the tables we created, and add <math>1</math>. We have
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| − | <cmath>D(3) = 1</cmath>
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| − | <cmath>D(6) = 3</cmath>
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| − | <cmath>D(12) = 8</cmath>
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| − | <cmath>D(24) = 20</cmath>
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| − | <cmath>D(48) = 48</cmath>
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| − | <cmath>D(96) = \boxed{\textbf{(A) }112}</cmath>
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| − | ==Solution 2==
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| − | Bash.
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| − | Since <math>96=2^5\times 3</math>, for the number of <math>f_n</math>, we have the following cases:
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| − | Case 1: <math>n=1</math>, we have <math>\{f_1\}=\{96\}</math>, only 1 case.
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| − | Case 2: <math>n=2</math>, we have <math>\{f_1,f_2\}=\{3,2^5\}, \{6,2^4\},...,\{48,2\}</math>, totally <math>5\cdot 2!=10</math> cases.
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| − | Case 3: <math>n=3</math>, we have <math>\{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\}</math>, totally <math>\frac{3!}{2!}\cdot 2+4\cdot 3!=30</math> cases.
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| − | Case 4: <math>n=4</math>, we have <math>\{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\}</math>, totally <math>\frac{4!}{2!}\cdot 3+\frac{4!}{3!}=40</math> cases.
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| − | Case 5: <math>n=5</math>, we have <math>\{f_1,f_2,f_3,f_4,f_5\}=\{3,2^2,2,2,2\},\{6,2,2,2,2\}</math>, totally <math>\frac{5!}{3!}+\frac{5!}{4!}=25</math> cases.
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| − | Case 6: <math>n=6</math>, we have <math>\{f_1,f_2,f_3,f_4,f_5,f_6\}=\{3,2,2,2,2,2\}</math>, totally <math>\frac{6!}{5!}=6</math> cases.
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| − | Thus, add all of them together, we have <math>1+10+30+40+25+6=112</math> cases. Put <math>\boxed{\textbf{(A) }112}</math>.
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| − | ~FANYUCHEN20020715
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| − | ==See Also==
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| − | {{AMC12 box|year=2020|ab=B|num-b=23|num-a=25}}
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| − | {{MAA Notice}}
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