How Many Friends Are On The Riddler Social Network?

A FiveThirtyEight Riddler puzzle.

By Vamshi Jandhyala in Riddler mathematics

November 5, 2021

Riddler Riddler Express

A group of $101$ people join $μετα$, and each person has a random, $50$ percent chance of being friends with each of the other $100$ people. Friendship is a symmetric relationship on $μετα$, so if you’re friends with me, then I am also friends with you.I pick a random person among the $101$ — let’s suppose her name is Marcia. On average, how many friends would you expect each of Marcia’s friends to have?

Computational Solution

From the simulation below, we see that the expected number of friends of each of Marcia’s friends is $\textbf{50.5}$. It is interesting to note that Marcia on average would have only $50$ friends.

import networkx as nx
from random import random, randint, choice

def exp_num_friends(n, runs = 10000):
    total_deg = 0
    for _ in range(runs):
        G = nx.Graph()
        for i in range(n):
        for i in range(n-1):
            for j in range(i+1, n):
                if random() < 0.5:
        marcia = randint(0,n-1)
        marcia_friends =  list(G.adj[marcia].keys())
        if marcia_friends:
            total_deg +=[choice(marcia_friends)]
    return total_deg/runs


Riddler Classic

The sum of the factors of $36$ — including $36$ itself — is $91$. Coincidentally, $36$ inches rounded to the nearest centimeter is … $91$ centimeters!

Can you find another whole number like $36$, where you can “compute” the sum of its factors by converting from inches to centimeters?

Extra credit: Can you find a third whole number with this property? How many more whole numbers can you find?

Computational Solution

From the code below we see that $378$ and $49600$ are two numbers below $1000000$ that satisify the given property.

from functools import reduce

def sum_of_factors(n):    
    return sum(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))

def inches_to_cm_same_as_sum_of_divisors():
    nums= []
    for i in range(37,1000000):
        if round(i*2.54) == sum_of_factors(i):
    return nums