Question

Let $S = \{(a, b) : a , b, \in Z, 0 \le a, a , b \le 18\}$. The number of elements $(x, y)$ in $S$ such that $3x + 4y + 5$ is divisible by $19$, is

• A. 38
• B. 19
• C. 18
• D. 1

Answer 1

Answer: (B)

We have,
$S = \{(a, b) : a, b, \in Z, 0 \le a, b \le 18\}$
$3x + 4y + 5 (x, y) \in S$
$\therefore S \le 3x + 4y + 5 \le 131$
[$\because$ min $(x, y) = (0, 0)$, max $(x, y) = (15,18)$]
Given, $3x + 4y + 5$ is divisible by $19$
$\therefore 3x + 4y + 5 = 19, 38, 57, 76, 95, 114 $
Case I
$3x + 4y + 5 = 19$
$ 3x + 4y = 14$
Only $(2, 2) $ satisfies
Case II
$3x + 4y + 5 = 38$
$3x + 4y = 33 $
Possible values of $ (x, y)$ is $(3, 6), (7, 3), (11, 0)$
Case III
$3x + 4y + 5 = 57$
$3x + 4y = 52$
Possible values of $ (x, y)$ is $(0, 13), (4, 10)$,
$(8, 7), (12, 4), (16, 1)$
Case IV
$3x + 4 y + 5=76$
Possible values of $(x, y)$ is $(1, 17), (5, 14)$,
$(9, 11), (13, 8), (17, 5)$
Case V
$3x + 4 y + 5 = 95$
Possible values of $(x, y)$ is
$(6,18), (10,15), (14,12), (18, 9)$
Case VI
$3x + 4 y + 5 = 114$
$(x, y) = (15,15)$
$\therefore $ Total solution $ = 19$