Difference between revisions of "2020 AMC 12B Problems/Problem 24"
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− | + | Bash. | |
+ | Since <math>96=2^5\times 3</math>, for the number of <math>f_n</math>, we have the following cases: | ||
+ | |||
+ | Case 1: <math>n=1</math>, we have <math>\{f_1\}=\{96\}</math>, only 1 case. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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{A}</math>. | ||
+ | |||
+ | ~FANYUCHEN20020715 | ||
{{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 01:22, 8 February 2020
Let denote the number of ways of writing the positive integer as a productwhere , the are integers strictly greater than , 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 can be written as , , and , so . What is ?
Solution
Bash. Since , for the number of , we have the following cases:
Case 1: , we have , only 1 case.
Case 2: , we have , totally cases.
Case 3: , we have , totally cases.
Case 4: , we have , totally cases.
Case 5: , we have , totally cases.
Case 6: , we have , totally cases.
Thus, add all of them together, we have cases. Put .
~FANYUCHEN20020715
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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