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Difference between revisions of "2021 JMC 10 Problems/Problem 16"

(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...")
 
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==Solution==
 
==Solution==
  
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We use an area-based approach along with number lines, where are total area is <math>10 \cdot 10=100</math>. We can stop worrying about what happens when at least one of <math>a</math> and <math>b</math> is an integer, because those cases are negligible. Below is the number line representing the value of <math>\lfloor |a| \rfloor</math> in the intervals. If seeing what happens on <math>a \in (n,n+1)</math> is too complicated, plugging in <math>a=n+\frac{1}{2}</math> will suffice.
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We use an area-based approach along with number lines, where are total area is <math>10 \cdot 10=100</math>. We can stop worrying about what happens when at least one of <math>a</math> and <math>b</math> is an integer, because those cases are negligible. Below is the number line representing the value of <math>\lfloor |a| \rfloor</math> in the intervals. If seeing what happens on <math>a \in (n,n+1)</math> is too complicated, plugging in <math>a=n+\tfrac{1}{2}</math> will suffice.
  
 
<asy>
 
<asy>
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We seek the number of <math>10 \cdot 10</math> pairs of <math>a</math> and <math>b</math> intervals where <math>\lfloor |a| \rfloor =| \lfloor b \rfloor |</math>. Doing simple casework on the equal value, we have <math>2 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 + 2 \cdot 2 + 2 \cdot 2 + 0 \cdot 1=18</math> cases, so our answer is <math>\frac{18}{10 \cdot 10}=\frac{9}{50}</math>.
+
We seek the number of <math>10 \cdot 10</math> pairs of <math>a</math> and <math>b</math> intervals where <math>\lfloor |a| \rfloor =| \lfloor b \rfloor |</math>. Doing simple casework on the equal value, we have <math>2 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 + 2 \cdot 2 + 2 \cdot 2 + 0 \cdot 1=18</math> cases, so our answer is <math>\tfrac{18}{10 \cdot 10}=\tfrac{9}{50}</math>.

Latest revision as of 15:32, 1 April 2021

Problem

If $a$ and $b$ are randomly chosen numbers between $-5$ and $5$, what is the probability that $\lfloor |a| \rfloor - | \lfloor b \rfloor |=0 ?$ (Recall that $\lfloor r \rfloor$ denotes the greatest integer less than or equal to $r.$)

$\textbf{(A) } \dfrac{1}{20} \qquad\textbf{(B) } \dfrac{2}{25} \qquad\textbf{(C) } \dfrac{1}{10} \qquad\textbf{(D) } \dfrac{9}{50} \qquad\textbf{(E) } \dfrac{1}{5}$

Solution

We use an area-based approach along with number lines, where are total area is $10 \cdot 10=100$. We can stop worrying about what happens when at least one of $a$ and $b$ is an integer, because those cases are negligible. Below is the number line representing the value of $\lfloor |a| \rfloor$ in the intervals. If seeing what happens on $a \in (n,n+1)$ is too complicated, plugging in $a=n+\tfrac{1}{2}$ 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]


Below is the number line representing the value of $| \lfloor b \rfloor |$ 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]


We seek the number of $10 \cdot 10$ pairs of $a$ and $b$ intervals where $\lfloor |a| \rfloor =| \lfloor b \rfloor |$. Doing simple casework on the equal value, we have $2 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 + 2 \cdot 2 + 2 \cdot 2 + 0 \cdot 1=18$ cases, so our answer is $\tfrac{18}{10 \cdot 10}=\tfrac{9}{50}$.

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