## A game is played with many identical, round coins on a flat table.

Consider a line perpendicular to the table.

The first coin is placed on the table touching the line.

Then, one by one, the coins are placed horizontally on top of the previous coin and touching the line.

The complete stack of coins must be balanced after every placement.

The diagram below [not to scale] shows a possible placement of 8 coins where point $P$ represents the line.

It is found that a minimum of $31$ coins are needed to form a coin loop around the line, i.e. if in the projection of the coins on the table the centre of the $n$th coin is rotated $\theta_n$, about the line, from the centre of the $(n-1)$th coin then the sum of $\displaystyle\sum_{k=2}^n \theta_k$ is first larger than $360^\circ$ when $n=31$. In general, to loop $k$ times, $n$ is the smallest number for which the sum is greater than $360^\circ k$.

Also, $154$ coins are needed to loop two times around the line, and $6947$ coins to loop ten times.

Calculate the number of coins needed to loop $2020$ times around the line.

### The problem could be solved by using the concept of geometric progression and harmonic series.

Given that a loop of k times about the line corresponds to sum of the angles being 360°k.

Since each coin is placed identically, the angle θn of the nth coin from the table base is a constant.

So θn = θ for all n. Therefore, for a circular loop around the line, θ1 + θ2 + θ3 … + θn = 360°k

which means n*θ = 360°k, and solving for θ, we arrive at θ = 360°k/n.

As more coins are added, the increment of θ values decreases. For a complete loop of k times around the line, n should be the smallest number such that θ1 + θ2 … + θn > 360°k, or in other words, θ1/n + θ2/n + … + θn/n > 360°k.

This resembles the concept of the Harmonic series: Hn = 1 + 1/2 + 1/3 + … + 1/n > ln(n) + γ, where γ is the Euler-Mascheroni constant, approximately 0.577.

Applying this to our equation, n*(θ1/n + θ2/n … + θn/n) > 360°k, gives us:

360°k < θ1 + θ2 + ... + θn = Hn * θ --> k < Hn * n / 360°.
Given that when k = 2, n = 154 and when k = 10, n = 6947, we can establish that n = Ce^(kH), where C ~ 154/e^2 is a proportionality constant and H ~ ln(6947/154)/8 is the harmonic constant.
Plugging k = 2020 into the equation results in n = Ce^(2020H), which becomes n = 154e^(2020 ln(6947/154)/8 - 2).
Evaluating the expression gives n approximately equal to 1.611*10^868, which is the number of coins needed to loop 2020 times around the line.
The answer is a very large number and clearly unfeasible in real life, but it accurately represents the mathematical solution to the problem.

##### More Answers:

A Bit of PrimeDivisors of $2n^2$

Paths to Equality