Geoboard Shapes

A geoboard (of order $N$) is a square board with equally-spaced pins protruding from the surface, representing an integer point lattice for coordinates $0 \le x, y \le N$.
John begins with a pinless geoboard. Each position on the board is a hole that can be filled with a pin. John decides to generate a random integer between $1$ and $N+1$ (inclusive) for each hole in the geoboard. If the random integer is equal to $1$ for a given hole, then a pin is placed in that hole.
After John is finished generating numbers for all $(N+1)^2$ holes and placing any/all corresponding pins, he wraps a tight rubberband around the entire group of pins protruding from the board. Let $S$ represent the shape that is formed. $S$ can also be defined as the smallest convex shape that contains all the pins.

The above image depicts a sample layout for $N = 4$. The green markers indicate positions where pins have been placed, and the blue lines collectively represent the rubberband. For this particular arrangement, $S$ has an area of $6$. If there are fewer than three pins on the board (or if all pins are collinear), $S$ can be assumed to have zero area.
Let $E(N)$ be the expected area of $S$ given a geoboard of order $N$. For example, $E(1) = 0.18750$, $E(2) = 0.94335$, and $E(10) = 55.03013$ when rounded to five decimal places each.
Calculate $E(100)$ rounded to five decimal places.

To solve this problem, we can iterate through all possible values of $i$ from 1 to $n$ and calculate the digit sum of $i$ in base $b_1$ and base $b_2$. If the digit sums are equal, we add $i$ to our result.

In the code below, we define a helper function `digit_sum` that calculates the digit sum of a number in a given base. Using this function, we iterate through all numbers from 1 to $n$, calculate their digit sums in base $b_1$ and base $b_2$, and add them to our result if they are equal.

“`python
def digit_sum(number, base):
# Calculate the digit sum of a number in a given base
sum = 0
while number != 0:
sum += number % base
number //= base
return sum

def M(n, b1, b2):
result = 0
for i in range(1, n+1):
if digit_sum(i, b1) == digit_sum(i, b2):
result += i
return result

sum_M = 0
for k in range(3, 7):
for l in range(1, k-1):
sum_M += M(10**16, 2**k, 2**l)

last_16_digits = sum_M % 10**16
print(last_16_digits)
“`

The code calculates the sum of $M(n, b_1, b_2)$ for each combination of $k$ and $l$ within the specified ranges. Finally, it calculates the last 16 digits by taking the modulo of the result with $10^{16}$.

Please note that solving this problem with $n = 10^{16}$ may take a significant amount of time due to the large number of iterations. You may try a smaller value for $n$ first to verify the correctness of the solution and then proceed with the larger value.

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