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Problem 641

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A Long Row of Dice

Consider a row of $n$ dice all showing $1$.

First turn every second die,$(2,4,6,\dots )$, so that the number showing is increased by $1$. Then turn every third die. The sixth die will now show a $3$. Then turn every fourth die and so on until every $n$th die (only the last die) is turned. If the die to be turned is showing a $6$ then it is changed to show a $1$.

Let $f(n)$ be the number of dice that are showing a $1$ when the process finishes. You are given $f(100)=2$ and $f({10}^8)=69$.

Find $f({10}^{36})$.

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一长排骰子

将共$n$个骰子排成一排，均翻至$1$点朝上。

从头开始，每数到第二个骰子（也即第$2,4,6,\dots$个骰子），就将其翻动至点数加$1$。然后再从头开始，每数到第三个骰子，就将其翻动至点数加$1$；此时第六个骰子的点数应该是$3$点。如此循环，在最后一轮，每数到第$n$个骰子才将其翻动至点数加$1$（也就是说，只翻动最后一个骰子）。在每次翻动前，如果骰子当前是$6$点朝上，则翻动后变为$1$点朝上。

令$f(n)$表示在全部翻动完毕后$1$点朝上的骰子数目。已知$f(100)=2$， $f({10}^8)=69$。

求$f({10}^{36})$。

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