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nth Term of A.P

Let S=a, b:...

Question

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

38

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19

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18

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1

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Solution

The correct option is D $19$
$I = 3 x + 4 y + 5 5 \leq I \leq 131$

Maximum at $(18, 18)$ and minumum at $(0, 0)$

$I$ is divided by $19$

Case $(i)$

$3 x + 4 y + 5 = 19 3 x + 4 y = 14 y = \frac{14 - 3 x}{4}$

Now we have to choose a put value of $x$ from our given domain so that $y$ also lies in our given domain

$\Rightarrow x = 2$

Case $(i i)$

$3 x + 4 y + 5 = 38 3 x + 4 y = 33 y = \frac{33 - 3 x}{4} \Rightarrow x = 3, 7, 11$

Case $(i i i)$

$3 x + 4 y + 5 = 57 3 x + 4 y = 52 y = \frac{52 - 3 x}{4} \Rightarrow x = 0, 4, 8, 12, 16$

Case $(i v)$

$3 x + 4 y + 5 = 76 3 x + 4 y = 71 y = \frac{71 - 3 x}{4} \Rightarrow x = 1, 5, 9, 13, 17$

Case $(v)$

$3 x + 4 y + 5 = 95 3 x + 4 y = 90 y = \frac{90 - 3 x}{4} \Rightarrow x = 6, 10, 14, 18$

Case $(v i)$

$3 x + 4 y + 5 = 114 3 x + 4 y = 109 y = \frac{109 - 3 x}{4} \Rightarrow x = 15$

$19$ cases are possible

Hence, option $B$ is correct.

Suggest Corrections

Similar questions

S = {(a, b) : a, b ϵ Z, 0 \leq a, b \leq 18}

(x, y)

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