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Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 7"

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== Solution ==
 
== Solution ==
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To have a number divisible by <math>12</math>, it must be divisible by <math>3</math> and <math>4</math>.
  
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<math>\frac{7}{18}</math>
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Consider mod 3 of x:
 +
 
 +
<math>x \: {\equiv} \: 0 \: (mod \: 3)</math>
 +
 
 +
<math>2^A + 10^B\:{\equiv}\:0\:(mod\:3)</math>
 +
 
 +
<math>2^A + 1\:{\equiv}\:0\:(mod\:3)</math>
 +
 
 +
<math>2^A\:{\equiv}\:2\:(mod\:3)</math>
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 +
Here, since <math>2^A\:{\equiv}\:1 \: or \: 2 \: (mod \: 3)</math> for A is even and odd respectively
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<math>\therefore \: A</math> is odd
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 +
Consider mod 4 of x:
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 +
<math>x \: {\equiv} \: 0 \: (mod \: 4)</math>
 +
 
 +
<math>2^A + 10^B\:{\equiv}\:0\:(mod\:4)</math>
 +
 
 +
<math>2^A + 2^B\:{\equiv}\:0\:(mod\:4)</math>
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 +
<math>\because\:</math> we know A is odd, for <math>A=1</math>, taking <math>mod\:4</math> gives <math>2</math> as result; for <math>A>1</math>, taking <math>mod\:4</math> gives <math>0</math> as result, so we split the case for <math>A=1</math> and <math>A>1</math> here.
 +
 
 +
For <math>A=1</math>,
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<math>2 + 2^B\:{\equiv}\:0\:(mod\:4)</math>
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<math>2^B\:{\equiv}\:2\:(mod\:4)</math>
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<math>\therefore \: B=1</math>
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For <math>A>1</math>,
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<math>2^B\:{\equiv}\:0\:(mod\:4)</math>
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<math>\therefore \: B>1</math>
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<math>\therefore\:</math> Concluding our above conditions, we have <math>(A,B) = (1,1)</math> or <math>(A,B) \in (\{3,5,7,9,11\},\:\{2,3,4,5,6,7,8,9,10,11,12\})</math>
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 +
By counting the number of solutions, the required probability <math>=\frac{1\times1 + 5\times11}{12\times12} = \frac{56}{144} = \frac{7}{18}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 11:57, 26 September 2022

Problem

Let $x = 2^A + 10^B$ where $A$ and $B$ are randomly chosen with replacement from among the positive integers less than or equal to twelve. What is the probability that $x$ is a multiple of $12$?

Solution

To have a number divisible by $12$, it must be divisible by $3$ and $4$.

Consider mod 3 of x:

$x \: {\equiv} \: 0 \: (mod \: 3)$

$2^A + 10^B\:{\equiv}\:0\:(mod\:3)$

$2^A + 1\:{\equiv}\:0\:(mod\:3)$

$2^A\:{\equiv}\:2\:(mod\:3)$

Here, since $2^A\:{\equiv}\:1 \: or \: 2 \: (mod \: 3)$ for A is even and odd respectively

$\therefore \: A$ is odd

Consider mod 4 of x:

$x \: {\equiv} \: 0 \: (mod \: 4)$

$2^A + 10^B\:{\equiv}\:0\:(mod\:4)$

$2^A + 2^B\:{\equiv}\:0\:(mod\:4)$

$\because\:$ we know A is odd, for $A=1$, taking $mod\:4$ gives $2$ as result; for $A>1$, taking $mod\:4$ gives $0$ as result, so we split the case for $A=1$ and $A>1$ here.

For $A=1$,

$2 + 2^B\:{\equiv}\:0\:(mod\:4)$

$2^B\:{\equiv}\:2\:(mod\:4)$

$\therefore \: B=1$

For $A>1$,

$2^B\:{\equiv}\:0\:(mod\:4)$

$\therefore \: B>1$

$\therefore\:$ Concluding our above conditions, we have $(A,B) = (1,1)$ or $(A,B) \in (\{3,5,7,9,11\},\:\{2,3,4,5,6,7,8,9,10,11,12\})$

By counting the number of solutions, the required probability $=\frac{1\times1 + 5\times11}{12\times12} = \frac{56}{144} = \frac{7}{18}$

See also

2018 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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