Question

Let \(A\) be the set of first \(101\) terms of an A.P., whose first term is \(1\) and the common difference is \(5\), and let \(B\) be the set of first \(71\) terms of an A.P., whose first term is \(9\) and the common difference is \(7\). Then the number of elements in \(A \cap B\), which are divisible by \(3\), is:

• A. \(4\)
• B. \(5\)
• C. \(6\)
• D. \(7\)

Answer 1

The Correct Option is B

Concept: A number common to two arithmetic progressions must satisfy both expressions. Such numbers can be found by solving a linear Diophantine equation. After obtaining the common sequence, we count those terms divisible by \(3\).
Step 1:Write the general terms of the two A.P.s.} For set \(A\): \[ a_k = 1 + (k-1)5 = 5k - 4, \qquad k=1,2,\dots,101 \] For set \(B\): \[ b_m = 9 + (m-1)7 = 7m + 2, \qquad m=1,2,\dots,71 \]
Step 2:Find common elements.} A common element must satisfy \[ 5k-4 = 7m+2 \] \[ 5k = 7m + 6 \] \[ k = \frac{7m+6}{5} \] For \(k\) to be integer, \[ 7m+6 \equiv 0 \pmod{5} \] \[ 2m+1 \equiv 0 \pmod{5} \] \[ m \equiv 2 \pmod{5} \] Thus \[ m = 2 + 5t \]
Step 3:Find the common sequence.} Substitute into \(b_m\): \[ b_m = 7(2+5t) + 2 \] \[ = 16 + 35t \] Thus the common elements form the A.P. \[ 16, 51, 86, 121, 156, 191, \dots \]
Step 4:Check bounds of the sets.} Largest element of \(A\): \[ 1 + 100(5) = 501 \] Largest element of \(B\): \[ 9 + 70(7) = 499 \] Thus common elements must be \(\le 499\). \[ 16 + 35t \le 499 \] \[ t \le 13 \]
Step 5:Count those divisible by \(3\).} \[ 16 + 35t \equiv 1 + 2t \pmod{3} \] For divisibility by \(3\): \[ 1 + 2t \equiv 0 \] \[ t \equiv 1 \pmod{3} \] Possible values within \(0 \le t \le 13\): \[ t = 1,4,7,10,13 \] Thus total numbers \(=5\).