Difference between revisions of "Mock AIME I 2015 Problems/Problem 1"
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| − | Associative Property! Note that the product of all four numbers can be written in two different ways, <math>(AB)(CD)</math> and <math>(AD)(BC)</math>. Setting these equal to each other gives <cmath>k=AD=\dfrac{(AB)(CD)}{BC}=\dfrac{14\times 18}{16}=\dfrac{ | + | Associative Property! Note that the product of all four numbers can be written in two different ways, <math>(AB)(CD)</math> and <math>(AD)(BC)</math>. Setting these equal to each other gives <cmath>k=AD=\dfrac{(AB)(CD)}{BC}=\dfrac{14\times 18}{16}=\dfrac{63}4\,\,\implies\,\, 20k=\boxed{315}.</cmath> |
Latest revision as of 21:21, 3 June 2017
Problem
David, Justin, Richard, and Palmer are demonstrating a "math magic" concept in front of an audience. There are four boxes, labeled A, B, C, and D, and each one contains a different number. First, David pulls out the numbers in boxes A and B and reports that their product is 
. Justin then claims that the product of the numbers in boxes B and C is 
, and Richard states the product of the numbers in boxes C and D to be 
. Finally, Palmer announces the product of the numbers in boxes D and A. If 
is the number that Palmer says, what is 
?
Solution
Associative Property! Note that the product of all four numbers can be written in two different ways, 
and 
. Setting these equal to each other gives ![\[k=AD=\dfrac{(AB)(CD)}{BC}=\dfrac{14\times 18}{16}=\dfrac{63}4\,\,\implies\,\, 20k=\boxed{315}.\]](http://latex.artofproblemsolving.com/b/c/a/bca4ee8c7ba06f40e8bda875a22e8177c934ce1f.png)
