2021 JMC 10 Problems/Problem 16
Revision as of 15:31, 1 April 2021 by Skyscraper (talk | contribs) (Created page with "==Problem== If <math>a</math> and <math>b</math> are randomly chosen numbers between <math>-5</math> and <math>5</math>, what is the probability that <math>\lfloor |a| \rfloo...")
Problem
If 
and 
are randomly chosen numbers between 
and 
, what is the probability that 
(Recall that 
denotes the greatest integer less than or equal to 
)
Solution
We use an area-based approach along with number lines, where are total area is 
. We can stop worrying about what happens when at least one of and is an integer, because those cases are negligible. Below is the number line representing the value of 
in the intervals. If seeing what happens on 
is too complicated, plugging in 
will suffice.
![[asy] import olympiad; size(4.5cm); int start=-5; int end=5; for(int i=start; i<end+1; ++i) { add(pathticks((i+1,0)--(i-1,0),1)); } draw((start-1,0)--(end+1,0),arrow=ArcArrows(HookHead)); label("$4$",(-4.5,0.3),N); label("$3$",(-3.5,0.3),N); label("$2$",(-2.5,0.3),N); label("$1$",(-1.5,0.3),N); label("$0$",(-0.5,0.3),N); label("$4$",(4.5,0.3),N); label("$3$",(3.5,0.3),N); label("$2$",(2.5,0.3),N); label("$1$",(1.5,0.3),N); label("$0$",(0.5,0.3),N); [/asy]](http://latex.artofproblemsolving.com/d/8/3/d83e1cafe31858ae3a764cae163a21a818e9fe52.png)
Below is the number line representing the value of 
in the intervals.
![[asy] import olympiad; size(4.5cm); int start=-5; int end=5; for(int i=start; i<end+1; ++i) { add(pathticks((i+1,0)--(i-1,0),1)); } draw((start-1,0)--(end+1,0),arrow=ArcArrows(HookHead)); label("$5$",(-4.5,0.3),N); label("$4$",(-3.5,0.3),N); label("$3$",(-2.5,0.3),N); label("$2$",(-1.5,0.3),N); label("$1$",(-0.5,0.3),N); label("$4$",(4.5,0.3),N); label("$3$",(3.5,0.3),N); label("$2$",(2.5,0.3),N); label("$1$",(1.5,0.3),N); label("$0$",(0.5,0.3),N); [/asy]](http://latex.artofproblemsolving.com/b/f/e/bfe7d439cafe71bce5aea97d461c604f730b424c.png)
We seek the number of 
pairs of and intervals where 
. Doing simple casework on the equal value, we have 
cases, so our answer is 
.
