Consecutive integers 8p, 9q, 10r: search for small solution sets (p,q,r) other than (1,1,1).

Question

The context: I want to show how prime factorisation patterns repeat along the number line. I've taken three consecutive integers, the first containing 2^3, the second 3^2, the third 5^1. Taking respective multiples p, q, r, this gives me a set of equations and a set of conditions, namely neither 2 nor 3 nor 5 can divide any of them. I've also found that p, q, r must all end in a '1'. What's the most efficient way to search for a solution set other than (1, 1, 1)?,

Answer 1

If $9|(8p+1)$ then you can prove that $ 8p\equiv -1(mod9)$ so $p\equiv 1 (mod9)$.

Similarly if $10|(9q+1)$ then try and prove that $q\equiv 1(mod 10)$.

If $10|(8p+2)$ $\implies 5|(4p+1)$ then $p\equiv 1(mod 5)$. Since 9, 5 are coprime, you can write that $p\equiv 1(mod 45)$. Hence, $$p=45x+1, q=40x+1, r=36x+1$$ form a solution for natural x. I derived q and r via $q=\frac{8p+1}{9}$ and $r=\frac{9q+1}{10}$.