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
$3x+4y+5=76$
Possible values of $(x,y)$ is $(1,17),(5,14)$,
$(9,11),(13,8),(17,5)$
Case V
$3x+4y+5=95$
Possible values of $(x,y)$ is
$(6,18),(10,15),(14,12),(18,9)$
Case VI
$3x+4y+5=114$
$(x,y)=(15,15)$
$\therefore$ Total solution $=19$