Difference between revisions of "1985 AIME Problems/Problem 1"
m (box) |
(→Solution 2) |
||
(7 intermediate revisions by 6 users not shown) | |||
Line 1: | Line 1: | ||
==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>. |
==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 < | + | 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! |
− | ==See also== | + | ==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 box|year=1985|before=First Question|num-a=2}} | ||
+ | * [[AIME Problems and Solutions]] | ||
+ | * [[American Invitational Mathematics Examination]] | ||
+ | * [[Mathematics competition resources]] | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Latest revision as of 15:07, 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 |