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Problem 641


Problem 641


A Long Row of Dice

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

First turn every second die,$(2,4,6,\ldots)$, 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 $f(10^{36})$.


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$2,4,6,\ldots$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$f(10^{36})$。