Tuesday, Jan 17th

Problem
Find the least positive integer $n$ such that $19/(n+21), 20/(n+22), \dots, 91/(n+93)$ are all irreducible fractions.

Solution
For all these fractions to be irreducible, we must have that $gcd(18+m, n+20+m)=1$ for $m=1,2,\dots, 73$.  Therefore, $gcd(18+m, n+2)=1$ which gives that $n+2$ has to have a prime divisor bigger than 18+73, thus $n+2=97$ is the smaller solution and $n=95$ gives the answer.