+826 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.
+1926 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! :)
