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If $m$ and $n$ are positive integers randomly chosen from the set $\{1, 2, \dots , 600\}$ with replacement, what is the probability that $2^m + 3^n$ is divisible by 11? Express your answer as a common fraction.

 Jun 5, 2024
 #1
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First, let's note that

\(2^3 + 3^1 = 11, 2^3 + 3^6 = 737 = 11 \cdot {67} \\ 2^3 + 3^{11} = 177155 = 16105 \cdot 11\)

 

Now, from this, we can tell that

\(m = 10a + 3 \\ n = 5b + 1\)

 

where a and b are different numbers. 

This means m has \( 3, 13, 23, ..., 593 = 60 \) values

And n has \(1, 6, 11, ..., 596 = 120 \) values. 

 

This means that there are \(60 * 120 = 7,200 \) cases that work.

In total, there are \(600 * 600 = 360,000\) different cases. 

 

Now, we just have \(7200/360000 = 72/3600 = 2/100 = 1/50\)

This is a 2% chance. 

 

Thanks! :)

 Jun 5, 2024

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