Common first term and difference in 3 arithmetic series

Question

Let $AP(a,d)$ denote the set of all the term of an infinite arithmetic progression with first term a and common difference d>0. If $AP(1,3)\ \cap AP(2,5)\ \cap AP(3,7)=AP(a,d)$ then find the value of a+d.

My approach

$1 + \left( {{n_1} - 1} \right)3 = X$

$2 + \left( {{n_2} - 1} \right)5 = X$

$3 + \left( {{n_3} - 1} \right)7 = X$

$1 + \left( {{n_1} - 1} \right)3 = 2 + \left( {{n_2} - 1} \right)5 = 3 + \left( {{n_3} - 1} \right)7 \Rightarrow 3{n_1} - 2 = 5{n_2} - 3 = 7{n_3} - 4 = X$

${n_1} = \frac{{X + 2}}{3};{n_2} = \frac{{X + 3}}{5};{n_3} = \frac{{X + 4}}{7}$

We need to find the value where $n_1, n_2,n_3$ are integer and through excel I found the first term as 52.

How do we calculate the first term a=52, d=LCM(3,5,7)=105

Answer 1

$n_1,n_2,n_3$ must be integers, which only happens when $$X=3p+1=5q+2=7r+3$$ for $p,q,r \in \mathbb Z$. This means $$3p=5q+1=7r+2 $$ The easiest way to proceed from here is to consider $7r+2$,it needs to be a multiple of $3$ and one more than a multiple of $5$. Since $7r+2 \equiv r+2 \equiv 0 \pmod 3$, you need $r\equiv 1 \pmod 3$. Then it is quick and easy to check $r=1,4,7$ and $7$ is the smallest $r$ that meets all the conditions. So, $$a=X=7(7)+3 =52$$