Using Continued fractions and having a batting average of .3085443, guess how many at-bats the player had.
Any tips or solutions so I can solve this problem? Currently I am reading about continued fractions, but I don't see the correct approach to this problem.
One of the essential facts about continued fractions is that their convergents give the best possible rational approximations to irrational numbers; also, the best possible approximations with small denominators to a rational number with a large denominator. The very best approximations are found by truncating the continued fraction just before a particularly large partial quotient.
So, the batting average is $$0.3085443=\frac{3085443}{10000000}\ .$$ If we take this literally, the player will have had $3085443$ successes from $10000000$ times at bat (or perhaps even $6170886$ from $20000000$, etc). This doesn't seem likely. So we assume that the number of times at bat was actually smaller, and the given decimal is a rounded number. We use continued fractions to work out the details. We can calculate $$0.3085443=\frac1{3+\displaystyle\frac1{4+\displaystyle\frac1{6+\displaystyle\frac1{1+\displaystyle\frac1{2+\displaystyle\frac1{1+\displaystyle\frac1{1+\displaystyle\frac1{658+\cdots}}}}}}}}$$ The first convergent is $\frac13$, meaning one success from three times at bat. This is clearly not right as the average would have been rounded to $0.3333333$ not $0.3085443$. The next convergent is $$\frac1{3+\displaystyle\frac1{4}}=\frac4{13}=0.3076923\cdots$$ which is wrong for the same reason. Keep trying until you find one that works. I haven't gone any further but my guess would be that the answer is $$\frac1{3+\displaystyle\frac1{4+\displaystyle\frac1{6+\displaystyle\frac1{1+\displaystyle\frac1{2+\displaystyle\frac1{1+\displaystyle\frac1{1}}}}}}}$$ found by truncating the continued fraction just before a particularly large partial quotient.