Set Definitions

Set A contains the first 101 terms of an arithmetic progression (A.P.) with first term $a_1 = 1$ and common difference $d_1 = 5$. The general term is $T_A(n) = a_1 + (n-1)d_1 = 1 + (n-1)5 = 5n - 4$, for $1 \le n \le 101$. The maximum value in A is $T_A(101) = 5(101) - 4 = 505 - 4 = 501$.

Set B contains the first 71 terms of an A.P. with first term $a'_1 = 9$ and common difference $d'_2 = 7$. The general term is $T_B(m) = a'_1 + (m-1)d'_2 = 9 + (m-1)7 = 7m + 2$, for $1 \le m \le 71$. The maximum value in B is $T_B(71) = 7(71) + 2 = 497 + 2 = 499$.

Terms Divisible by 3 in Set A

A term $T_A(n) = 5n - 4$ is divisible by 3 if:

$5n - 4 \equiv 0 \pmod{3}$

$2n - 1 \equiv 0 \pmod{3}$

$2n \equiv 1 \pmod{3}$

Multiplying by 2 (the inverse of 2 mod 3):

$4n \equiv 2 \pmod{3}$

$n \equiv 2 \pmod{3}$

So, terms in A divisible by 3 occur when $n$ is of the form $3k - 1$ (for $k=1, 2, ...$). Let's check the range: $n = 2, 5, 8, ...$. These terms form an A.P. $T_A(2) = 5(2)-4 = 6$. The common difference for this subsequence is $3 \times d_1 = 3 \times 5 = 15$. The sequence is $6, 21, 36, ...$.

Terms Divisible by 3 in Set B

A term $T_B(m) = 7m + 2$ is divisible by 3 if:

$7m + 2 \equiv 0 \pmod{3}$

$m + 2 \equiv 0 \pmod{3}$

$m \equiv -2 \equiv 1 \pmod{3}$

So, terms in B divisible by 3 occur when $m$ is of the form $3j - 2$ (for $j=1, 2, ...$). Let's check the range: $m = 1, 4, 7, ...$. These terms form an A.P. $T_B(1) = 7(1)+2 = 9$. The common difference for this subsequence is $3 \times d'_2 = 3 \times 7 = 21$. The sequence is $9, 30, 51, ...$.

Intersection of Sets A and B Divisible by 3

We need to find the common terms in the two subsequences derived above:

• Subsequence from A (divisible by 3): $6, 21, 36, 51, 66, 81, 96, ...$ (A.P. with first term 6, common difference 15).
• Subsequence from B (divisible by 3): $9, 30, 51, 72, 93, 114, ...$ (A.P. with first term 9, common difference 21).

The common terms will form an A.P. with a common difference equal to the LCM of 15 and 21.

$LCM(15, 21) = LCM(3 \times 5, 3 \times 7) = 3 \times 5 \times 7 = 105$.

The first common term is 51.

The sequence of common terms (in $A \cap B$ and divisible by 3) starts with 51 and has a common difference of 105. The general term is $51 + k \times 105$, where $k \ge 0$.

Counting Common Elements

These common terms must be within the bounds of both Set A and Set B. The maximum possible value is $\min(501, 499) = 499$.

We need to find the number of terms $k$ such that:

$51 + k \times 105 \le 499$

$k \times 105 \le 499 - 51$

$k \times 105 \le 448$

$k \le \frac{448}{105} \approx 4.267$

Since $k$ must be an integer, the maximum value for $k$ is 4.

The possible values for $k$ are $0, 1, 2, 3, 4$. This gives a total of $4 + 1 = 5$ terms.

The terms are: 51, 156, 261, 366, 471.

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