What is the probability that gcd(n,m)=1?  (from the Prime Pages' list of frequently asked questions)
 New record prime: 277,232,917-1 with 23,249,425 digits by Pace, Woltman, Kurowski, Blosser & GIMPS (26 Dec 2017).
 What is the probability that gcd(n,m)=1? Here is a frequently asked question:  While playing with programs to determine primes and relative primes, I stumbled over an interesting (at least to me) fact. While the probability of a random number being prime decreases as the range of possible random numbers increases (Prime Number Theorem), the probability of two random numbers being relatively prime is 60.8% Is this something that is either well known by or trivially obvious to prime number gurus?  That there should be such a constant is "obvious", finding its value takes more work. In 1849 Dirichlet showed that the probability is 6/2 roughly as follows.  Suppose you pick two random numbers less than n, then  [n/2]2 pairs are both divisible by 2.  [n/3]2 pairs are both divisible by 3.  [n/5]2 pairs are both divisible by 5.  ...  (Here [x] is the greatest integer less than or equal to x, usually called the floor function.) So the number of relatively prime pairs less than or equal to n is (by the inclusion/exclusion principle):  n2 - sum([n/p]2) + sum([n/pq]2) - sum([n/pqr]2) + ...  where the sums are taken over the primes p,q,r,... less than n. Letting mu(x) be the möbius function this is  sum(mu(k)[n/k]2)     (sum over positive integers k)  so the desired constant is the limit as n goes to infinity of this sum divided by n2, or  sum(mu(k)/k2)     (sum over positive integers k).  But this series times the sum of the reciprocals of the squares is one, so the sum of this series, the desired limit, is 6/2. This number is approximately  60.7927101854026628663276779258365833426152648033479  percent.  ;-)
 Another prime page by Chris K. Caldwell