Difference between revisions of "1985 AIME Problems/Problem 1"
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==Problem== | ==Problem== | ||
− | Let <math>x_1=97</math>, and for <math>n>1</math> let<math>x_n=\frac{n}{x_{n-1}}</math>. Calculate the product <math>x_1x_2x_3x_4x_5x_6x_7x_8</math>. | + | Let <math>x_1=97</math>, and for <math>n>1</math>, let <math>x_n=\frac{n}{x_{n-1}}</math>. Calculate the [[product]] <math>x_1x_2x_3x_4x_5x_6x_7x_8</math>. |
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==Solution== | ==Solution== | ||
− | --- | + | Since <math>x_n=\frac{n}{x_{n-1}}</math>, <math>x_n \cdot x_{n - 1} = n</math>. Setting <math>n = 2, 4, 6</math> and <math>8</math> in this equation gives us respectively <math>x_1x_2 = 2</math>, <math>x_3x_4 = 4</math>, <math>x_5x_6 = 6</math> and <math>x_7x_8 = 8</math> so <cmath>x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.</cmath> Notice that the value of <math>x_1</math> was completely unneeded! |
− | *[[ | + | |
+ | ==Solution 2== | ||
+ | Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel: | ||
+ | |||
+ | <cmath>x_1x_2x_3x_4x_5x_6x_7x_8=x_1\cdot\dfrac{2}{x_1}\cdot\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}\cdot\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}\cdot\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}}</cmath> | ||
+ | <cmath>=\left (x_1\cdot\dfrac{2}{x_1} \right )\cdot \left (\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}} \right )\cdot \left (\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}} \right )\cdot \left (\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}} \right )</cmath> | ||
+ | <cmath>=(2)\cdot (4)\cdot (6)\cdot (8)=\boxed{384}</cmath> | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=1985|before=First Question|num-a=2}} | ||
+ | * [[AIME Problems and Solutions]] | ||
+ | * [[American Invitational Mathematics Examination]] | ||
+ | * [[Mathematics competition resources]] | ||
− | + | [[Category:Introductory Algebra Problems]] |
Latest revision as of 14:07, 4 September 2020
Contents
Problem
Let , and for , let . Calculate the product .
Solution
Since , . Setting and in this equation gives us respectively , , and so Notice that the value of was completely unneeded!
Solution 2
Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel:
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |