If ${[3 (230 + x)]}^{2} = 492 a 04$ , then $a + x$ is

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Solution

$[3 (230 + x)]^{2} = 492 a 04$
$(230 + x)^{2} = \frac{492 a 04}{9}$
For $492 a 04$ to be divisible by $9$ , sum of digits must be divisible by $9$
$⟹ 4 + 9 + 2 + a + 0 + 4 = 19 + a ⟹ a = 8$
$3 (230 + x) = \sqrt{492804} = 702 ⟹ x = 4$
$a + x = 12$

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Similar questions

{(3 (230 + x))^{2}} = 492 a 04

a

a = \frac{4}{3} {cot}^{2} 30^{\circ} + 3 {sin}^{2} 60^{\circ} - 3 {sec}^{2} 30^{\circ} - \frac{3}{4} {tan}^{2} 30^{\circ}

b = 3 {tan}^{2} 45^{\circ} + cos 0^{\circ} - cot 90^{\circ}

{log}_{b} (a) =

\frac{4}{3} {cot}^{2} 30^{\circ} + 3 {sin}^{2} 60^{\circ} - 2 {cosec}^{2} 60^{\circ} - \frac{3}{4} {tan}^{2} 30^{\circ}

2 {tan}^{2} 60^{\circ} - 4 {cos}^{2} 45^{\circ} - 3 {sec}^{2} 30^{\circ}

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