Let A be the set of the first 101 terms of the A.P.

Question

Let $A$ be the set of the first $101$ terms of the A.P. with first term $1$ and common difference $5$, and let $B$ be the set of the first $71$ terms of the A.P. with first term $9$ and common difference $7$. Then the number of elements of $A \cap B$ that are divisible by $3$ is:

Accepted Answer

$A = \{1,6,11,\ldots,501\}$ (last term $a_{101}=1+100	imes5=501$).
$B = \{9,16,23,\ldots,499\}$ (last term $b_{71}=9+70	imes7=499$).

The common terms form an A.P. starting at $16$ with common difference $\mathrm{lcm}(5,7)=35$: $C_k = 16+35(k-1)$.

Requiring $C_k \le \min(501,499)=499$: $35(k-1)\le483 \Rightarrow k-1\le13.8 \Rightarrow k\le14$. So $A\cap B$ has $14$ terms.

For divisibility by $3$: $16\equiv1$, $35\equiv2 \pmod3$, so $C_k \equiv 1+2(k-1) \pmod 3$. This is $\equiv0\pmod3$ when $k-1\equiv1\pmod3$, i.e. $k \in \{2,5,8,11,14\}$ --- exactly $5$ values.

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