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Difference between revisions of "2021 April MIMC 10 Problems/Problem 14"

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<math>\textbf{(A)} ~\frac{1}{25} \qquad\textbf{(B)} ~\frac{2}{45} \qquad\textbf{(C)} ~\frac{11}{225} \qquad\textbf{(D)} ~\frac{4}{75} \qquad\textbf{(E)} ~\frac{13}{225}</math>
 
<math>\textbf{(A)} ~\frac{1}{25} \qquad\textbf{(B)} ~\frac{2}{45} \qquad\textbf{(C)} ~\frac{11}{225} \qquad\textbf{(D)} ~\frac{4}{75} \qquad\textbf{(E)} ~\frac{13}{225}</math>
 
==Solution==
 
==Solution==
−
We can begin by converting all the elements in the set to Modular of <math>5</math>. Then, we realize that all possible elements that can satisfy all the expressions to be divisible by <math>5</math> can only happen if <math>x</math> and <math>y</math> are both <math>0</math> (mod <math>5)</math>. Since <math>x</math> and <math>y</math> are not necessarily distinct, we have <math>3^2=9</math> possible <math>(x,y)</math>. There are total of <math>15\cdot 15=225</math> possible <math>(x,y)</math>, therefore, the probability is <math>\frac{9}{25}=\boxed{\textbf{(A)} \frac{1}{25}}</math>.
+
We can begin by converting all the elements in the set to Modular of <math>5</math>. Then, we realize that all possible elements that can satisfy all the expressions to be divisible by <math>5</math> can only happen if <math>x</math> and <math>y</math> are both <math>0</math> (mod <math>5)</math>. Since <math>x</math> and <math>y</math> are not necessarily distinct, we have <math>3^2=9</math> possible <math>(x,y)</math>. There are total of <math>15\cdot 15=225</math> possible <math>(x,y)</math>, therefore, the probability is <math>\frac{9}{225}=\boxed{\textbf{(A)} \frac{1}{25}}</math>.

Latest revision as of 01:05, 15 June 2021

James randomly choose an ordered pair $(x,y)$ which both $x$ and $y$ are elements in the set $\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$, $x$ and $y$ are not necessarily distinct, and all of the equations: \[x+y\] \[x^2+y^2\] \[x^4+y^4\] are divisible by $5$. Find the probability that James can do so.

$\textbf{(A)} ~\frac{1}{25} \qquad\textbf{(B)} ~\frac{2}{45} \qquad\textbf{(C)} ~\frac{11}{225} \qquad\textbf{(D)} ~\frac{4}{75} \qquad\textbf{(E)} ~\frac{13}{225}$

Solution

We can begin by converting all the elements in the set to Modular of $5$. Then, we realize that all possible elements that can satisfy all the expressions to be divisible by $5$ can only happen if $x$ and $y$ are both $0$ (mod $5)$. Since $x$ and $y$ are not necessarily distinct, we have $3^2=9$ possible $(x,y)$. There are total of $15\cdot 15=225$ possible $(x,y)$, therefore, the probability is $\frac{9}{225}=\boxed{\textbf{(A)} \frac{1}{25}}$.

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