# Batting average

How would you interpret this problem? *

Let $$h$$ and $$n$$ be the number of hits and number of at-bats after hitting the single, respectively. The answer to the problem suggests solving from the following identity:

$$\frac{h - 1}{n - 1} + \frac{1}{100} = \frac{h}{n}$$

I started from:

$$\lfloor \frac{h-1}{n - 1} \ 1000 \rfloor + 10 = \lfloor {\frac{h}{n} \ 1000} \rfloor$$

Which seems much harder to solve. I don't know how to do it in general. Computer search counted 336 pairs $$(h, n)$$ with $$h$$ between 2 and 30.

(*) "Which Way Did the Bicycle Go?", Kornhauser et al., 1996, p. 34

• I would understand the phrase "went up by exactly 10 points" to literally mean exactly--no rounding. This is also borne out in the 'Note' at the end of the problem statement. Commented Oct 15, 2018 at 12:47
• @paw88789 The "traditional rounding" steered me to this scenario: a player is just about to bat. TV says his batting average is 0.abc. After he bats, TV says his batting average is 0.def.
– BoLe
Commented Oct 15, 2018 at 13:01
• Given that the answer produces batting averages that are terminating decimals with less than three digits after the decimal point, I think the problem is unnecessarily confusing. As @paw88789 pointed out, there are multiple interpretations. Commented Oct 15, 2018 at 13:26
• I think the traditionally given is just there to justify the $10$ points. Your first equation, without the floors, is the intended one. Commented Oct 15, 2018 at 13:54

We have $$\frac{h - 1}{n - 1} + \frac{1}{100} = \frac{h}{n}\\100n(h-1)+n(n-1)=100h(n-1)\\ h=n-\frac {n(n-1)}{100}$$ So we need $$n(n-1)$$ to be a multiple of $$100$$. One of $$n,n-1$$ needs to be a multiple of $$25$$ and the other a multiple of $$4$$. This gives $$n=25, h=19$$
He started at $$\frac {18}{24}=0.750$$ and ended at $$\frac {19}{25}=0.760$$. Quite a hitter. The next solution has $$n=75$$ and it turns out that also works with $$18$$ hits, with the average going from $$0.240$$ to $$0.250$$. In either case he started with $$18$$ hits. Thanks to paw88789 for catching my error
• In fact, $18$ for $75$ gives a batting average of $.240$. And $19$ for $76$ is $.250$ Commented Oct 16, 2018 at 19:17