How would you interpret this problem? *

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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

  • 1
    $\begingroup$ 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. $\endgroup$
    – paw88789
    Commented Oct 15, 2018 at 12:47
  • $\begingroup$ @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. $\endgroup$
    – BoLe
    Commented Oct 15, 2018 at 13:01
  • 1
    $\begingroup$ 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. $\endgroup$
    – rogerl
    Commented Oct 15, 2018 at 13:26
  • 1
    $\begingroup$ I think the traditionally given is just there to justify the $10$ points. Your first equation, without the floors, is the intended one. $\endgroup$ Commented Oct 15, 2018 at 13:54

1 Answer 1


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

  • $\begingroup$ In fact, $18$ for $75$ gives a batting average of $.240$. And $19$ for $76$ is $.250$ $\endgroup$
    – paw88789
    Commented Oct 16, 2018 at 19:17
  • $\begingroup$ @paw88789: you are right. As the question just asked for the number of hits we have a unique answer $\endgroup$ Commented Oct 16, 2018 at 19:21

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