Question

Let $S=\left\{ \left( a,b \right):a,b\in Z,0\le a,b\le 18 \right\}$. The number of elements $\left( x,y \right)$ in S such that $3x+4y+5$ is divisible by 19 is?
a) 38
b) 19
c) 18
d) 1

Answer 1

Hint:We will find the upper and lower limit of the function $f\left( x \right)=3x+4y+5$ upper limit is 5 and lower limit is 131. $3x+4y+5$ is divisible by 19, it means 19 is the multiple of $3x+4y+5$. We can write it as - $3x+4y+5=19\times I$, where ‘I’ is any integral value between 0 and 18 as $0\le a,b\le 18$. Now, we will consider all the cases possible to find the number of elements in $\left( x,y \right)$ in S which are divisible by 19.

Complete step-by-step answer:
It is given in the question that $S=\left\{ \left( a,b \right):a,b\in Z,0\le a,b\le 18 \right\}$ then we have to find that the number of elements $\left( x,y \right)$ such that $3x+4y+5$ is divisible by 19.
It is clearly given that $3x+4y+5$ is divisible by 19 which means 19 is a factor of $3x+4y+5$ thus it can be written as - $3x+4y+5=19\times I$, where ‘I’ is any integral value. But, as per the conditions, we get that $3x+4y+5$ is limited in between $0\le a,b\le 18$. Thus, we get that $3x+4y+5$ is minimum at $\left( 0,0 \right)$ and maximum at $\left( 18,18 \right)$.
Now, we get the minimum value of $3x+4y+5$ at $\left( 0,0 \right)$ by substituting in it as $3\left( 0 \right)+4\left( 0 \right)+5=5$, also the maximum value of $3x+4y+5$ at $\left( 18,18 \right)$ by substituting in it as $3\left( 18 \right)+4\left( 18 \right)+5=131$. Therefore, we get $5\le (3x+4y+5)\le 131$.
Now, the multiples of 19 which are lying in between 5 and 131 are- 19,38,57,76,95,119. So there will be six cases possible-
Case-1 When $3x+4y+5=19$
$y=\dfrac{19-5-3x}{4}=\dfrac{14-3x}{4}$.
Now we will try to find the value of x such that it is from our given domain. Our domain is $0\le a,b\le 18$. Only one case is possible when $x=2$.

Case-2 When $3x+4y+5=19\times 2=38$
$y=\dfrac{38-5-3x}{4}=\dfrac{33-3x}{4}$,
Thus, we get the possible values of x = 3,7,11.

Case-3 When $3x+4y+5=19\times 3=57$,
$y=\dfrac{57-5-3x}{4}=\dfrac{52-3x}{4}$
Thus we get the possible values of x = 0,4,8,12,16.

Case-4 When $3x+4y+5=19\times 4=76$,
$y=\dfrac{76-5-3x}{4}=\dfrac{71-3x}{4}$
Thus we get the possible values of x = 1,5,9,13,17.

Case-5 When $3x+4y+5=19\times 5=95$,
$y=\dfrac{95-5-3x}{4}=\dfrac{90-3x}{4}$
Thus we get the possible values of x = 6,10,14,18.

Case-6 When $3x+4y+5=19\times 6=114$,
$y=\dfrac{114-5-3x}{4}=\dfrac{109-3x}{4}$
Thus we get a possible value of x = 15.
To get the total number of elements in $\left( x,y \right)$ in S we have to add all numbers obtained values of x in all six possible cases. So, we get- \[total\text{ }elements=1+3+5+3+6+1=19\].
Thus the total number of elements in $\left( x,y \right)$ in S will be 19. Therefore, option b) will be the correct answer.

Note: Students may start to find the whole pairs of elements possible in $\left( x,y \right)$ in S but it is not asked in the question, therefore it is not necessary to find all the pairs. It is always recommended to find only those which were asked in question.We have to substitute one by one values of x between the given condition of domain and also range of y should satisfies the condition, So form the pairs of elements possible for the cases of all multiples of 19 as given above.