Common terms of two A.P Series : Indian Statistical Institute B.Math & B.Stat

Consider the two arithmetic progressions $3,7,11,\dots,407$ and $2,9,16,\dots,709$. Find the number of common terms of these two progressions. Let $a_n$ and $a_m$ be the last terms of the progressions respectively $\Rightarrow 407 = 3+(n-1)4$ and $709 = 2+(m-1)7$ Solving we get, $n,m = 102$. To find the common terms, assume that the $n^{th}$ term of the first progression is equal to the $m^{th}$ term of the second progression. $\Rightarrow 3+(n-1)4=2+(m-1)7 \Rightarrow 3+4n-4-2+7=7m \Rightarrow 4(n+1)=7m,$ where $n,m \in \{1,2,3,\dots,102\}$ R.H.S is a multiple of 7, while L.H.S is 4(n+1). Since $g.c.d(4,7)=1$ L.H.S will be multiple of $7$, $iff$ $n+1$ is a multiple of $7$. $\Rightarrow n = 6,13,20,\dots$ Again since $n$ is bounded by $102$. The largest possible value of $n$ is $97$. So, $n \in \{6,13,20,\dots\,97}$ Which has $14$ terms. Thus the number of common terms of the progression is $14$