2.1. Let $A=\{3+10 k, 6+26 k, 5+29 k, k=1,2,3,4, \cdots\}$. Determine the smallest positive integer $r$ such that there exists an integer $b$ with the property that the set $B=\{b+r k, k=1,2,3,4, \cdots\}$ is disjoint from $A$
2.1 The answer is 290 . First observe that the arithmetic sequences $\{p+m k, k \in \mathbb{N}\}$ and $\{q+n k, k \in \mathbb{N}\}$ are disjoint if and only if $p-q \neq l n-k m$ for all integers $k, l$, which holds if and only if $\operatorname{gcd}(m, n)$ does not divide $p-q$. Therefore, the required $r$ cannot be relatively prime to $10=(2)(5), 26=(2)(13)$ and 29 . We start by choosing $r$ to be the smallest $\mathrm{lcm}$ of $d_{1}, d_{2}, d_{3}$ where $d_{1}, d_{2}, d_{3}$ are factors (greater than 1 ) of 10,26 and 29 respectively. The smallest such $r$ is $58=(2)(29)$. In this case, $\operatorname{gcd}(10,58)=2, \operatorname{gcd}(26,58)=2$ and $\operatorname{gcd}(29,58)=29$. We also require $b-3 \not \equiv 0(\bmod 2), b-6 \not \equiv 0(\bmod 2)$ and $b-5 \not \equiv 0$ (mod 29). But there is no solution for $b$ from the first two equations. Therefore we cannot take $r=58$. The next smallest lcm would be $290=(5)(2)(29)$. In this case, a simple checking using the above criterion shows that $\{1+290 k, k \in \mathbb{N}\}$ is disjoint from $A$