+1537 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.
+1632 2^3 + 3^1 = 11, 2^3 + 3^6 = 737 = 11*67, 2^3 + 3^11 = 177155 = 16105*11.
2^13 + 3^1 is divisible by 11, same as 2^23 + 3^1 is also. Along with 2^13 + 3^6...
As you will notice with these calculations (that you can do brutally by hand or easily with a calculator (to find patterns, of course)), the values of m and n for 2^m + 3^n is divisible by 11 is m = 10a + 3, n = 5b + 1.
List for m: 3, 13, 23, ..., 593 = 60 values
List for n: 1, 6, 11, ..., 596 = 120 values
In total, there are 60 * 120 = 7,200 correct cases, along with 600 * 600 = 360,000 total possible cases: coming to a probability of 7200/360000 = 72/3600 = 2/100 = 1/50, or 2% chance.
Now, I leave you with this parmen... how can you prove that the generalizations for the patterns of m and n are correct?? 
+1632 2^3 + 3^1 = 11, 2^3 + 3^6 = 737 = 11*67, 2^3 + 3^11 = 177155 = 16105*11.
2^13 + 3^1 is divisible by 11, same as 2^23 + 3^1 is also. Along with 2^13 + 3^6...
As you will notice with these calculations (that you can do brutally by hand or easily with a calculator (to find patterns, of course)), the values of m and n for 2^m + 3^n is divisible by 11 is m = 10a + 3, n = 5b + 1.
List for m: 3, 13, 23, ..., 593 = 60 values
List for n: 1, 6, 11, ..., 596 = 120 values
In total, there are 60 * 120 = 7,200 correct cases, along with 600 * 600 = 360,000 total possible cases: coming to a probability of 7200/360000 = 72/3600 = 2/100 = 1/50, or 2% chance.
Now, I leave you with this parmen... how can you prove that the generalizations for the patterns of m and n are correct??
