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) | + | 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>. |
Revision as of 12:44, 26 April 2021
James randomly choose an ordered pair 
which both 
and 
are elements in the set 
, and are not necessarily distinct, and all of the equations:
![\[x+y\]](http://latex.artofproblemsolving.com/1/f/8/1f8c15b4e27b092ee81e53929bf430612e47048c.png)
![\[x^2+y^2\]](http://latex.artofproblemsolving.com/a/f/e/afe2c766eb539f2173fc00ba60b373a430f553c8.png)
![\[x^4+y^4\]](http://latex.artofproblemsolving.com/4/6/1/461f2a5af17f9fc2f961808da11215a183252013.png)
are divisible by 
. Find the probability that James can do so.
Solution
We can begin by converting all the elements in the set to Modular of . Then, we realize that all possible elements that can satisfy all the expressions to be divisible by can only happen if and are both 
(mod 
. Since and are not necessarily distinct, we have 
possible . There are total of 
possible , therefore, the probability is $\frac{9}{25}=\boxed{\textbf{(A) \frac{1}{25}}}$ (Error compiling LaTeX. Unknown error_msg).
