Number of Elements Divisible by Three in Intersection of Progressions

To find the number of elements in $A \cap B$ that are divisible by $3$, we first determine the elements of the sets $A$ and $B$, find their intersection, and then count how many terms in $A \cap B$ are divisible by $3$.

Step 1: Elements of Set $A$

The set $A$ consists of the first $101$ terms of an arithmetic progression with first term $a_1 = 1$ and common difference $d_1 = 5$. The $n^{\text{th}}$ term of $A$ is given by: $T_n^{(A)} = 1 + (n - 1)5 = 5n - 4, \quad \text{for } 1 \le n \le 101$

The largest element in set $A$ is: $T_{101}^{(A)} = 1 + 100 \times 5 = 501$

Step 2: Elements of Set $B$

The set $B$ consists of the first $71$ terms of an arithmetic progression with first term $a_2 = 9$ and common difference $d_2 = 7$. The $m^{\text{th}}$ term of $B$ is given by: $T_m^{(B)} = 9 + (m - 1)7 = 7m + 2, \quad \text{for } 1 \le m \le 71$

The largest element in set $B$ is: $T_{71}^{(B)} = 9 + 70 \times 7 = 499$

Step 3: Intersection Set $A \cap B$

An element $x$ belongs to $A \cap B$ if it is a common term of both sequences. Equating the general terms: $5n - 4 = 7m + 2 \implies 5n - 7m = 6$

By inspection, the smallest common term occurs when $n = 4$ and $m = 2$: $x_1 = 5(4) - 4 = 7(2) + 2 = 16$

The common difference of the intersection progression is: $d = \text{lcm}(d_1, d_2) = \text{lcm}(5, 7) = 35$

Thus, the common terms form an A.P. given by: $C_k = 16 + (k - 1)35 = 35k - 19, \quad \text{for } k \ge 1$

Since any common element cannot exceed the maximum element of $B$ (which is $499$): $35k - 19 \le 499$ $35k \le 518$ $k \le \frac{518}{35} \approx 14.8$

Since $k$ must be a positive integer: $1 \le k \le 14$

Thus, there are $14$ elements in $A \cap B$.

Step 4: Finding Elements Divisible by 3

We need to find $k \in \{1, 2, \dots, 14\}$ such that $C_k$ is divisible by $3$: $C_k \equiv 0 \pmod 3$ $35k - 19 \equiv 0 \pmod 3$

Reducing modulo $3$: $(2)k - (1) \equiv 0 \pmod 3$ $2k \equiv 1 \pmod 3$

Multiplying both sides by $2$: $4k \equiv 2 \pmod 3 \implies k \equiv 2 \pmod 3$

Therefore, $k$ must be of the form $k = 3j + 2$ for non-negative integers $j$.

Since $1 \le k \le 14$, the valid values for $k$ are:

• For $j = 0$: $k = 2$
• For $j = 1$: $k = 5$
• For $j = 2$: $k = 8$
• For $j = 3$: $k = 11$
• For $j = 4$: $k = 14$

Thus, there are $5$ such values of $k$ (which yield the divisible elements: $51, 156, 261, 366, 471$).

Hence, the number of elements in $A \cap B$ which are divisible by $3$ is 5.

Correct Answer: B