Each of the numbers except 81 is a prime number.

23 + 2² = 27

27 + 3² = 36

36 + 4² = 52

52 + 5² = 77

77 + 6² = 113

113 + 7² = 162

Hence, 113 should have come in place of 111

24 = 3 x8, where 3 and 8 co-prime.

Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8.

Similarly, 63810 is not divisible by 8 and 537804 is not divisible by 8.

Consider option (D),

Sum of digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3.

Also, 736 is divisible by 8.

$\therefore$ 3125736 is divisible by 3 x 8, i.e., 24.

Required numbers are 24, 30, 36, 42, ..., 96

This is an A.P. in which $ a $ = 24, $ d $ = 6 and $ l $ = 96

Let the number of terms in it be $ n $.

Then t_n = 96  $\Rightarrow$  $ a $ + $\left(n - 1\right)$d = 96

$\Rightarrow$ 24 + $\left( n - 1\right)$ x 6 = 96

$\Rightarrow$ $\left( n - 1\right)$ x 6 = 72

$\Rightarrow$ $\left( n - 1\right)$ = 12

$\Rightarrow n $ = 13

Required number of numbers = 13.

Let S be the sample space.

Then, $ n \left(S\right)$ = ⁵²C₂ =$ \dfrac{(52 \times 51)}{(2 \times 1)} $= 1326.

Let E = event of getting 2 kings out of 4.

$\therefore n \left(E\right)$ = ⁴C₂ =$ \dfrac{(4 \times 3)}{(2 \times 1)} $= 6.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{6}{1326} $=$ \dfrac{1}{221} $.

$n\left(S\right)$ = Total number of ways of drawing 2 cards from 52 cards = ⁵²C₂

Let E = event of getting 1 club and 1 diamond.

We know that there are 13 clubs and 13 diamonds in the total 52 cards.

$Hence, n\left(E\right)$ = Number of ways of drawing one club from 13 and one diamond from 13

= ¹³C₁ × ¹³C₁

$\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} $= $\dfrac{13_{C_1} \times 13_{C_1}}{52_{C_2}}$

= $\dfrac{13 \times 13}{\left( \dfrac{52 \times 51}{2}\right)}$=$ \dfrac{13 \times 13}{ 26 \times 51}= \dfrac{13}{ 2\times 51}$=$ \dfrac{13}{102}$

22 – (1/4) {–5 – (– 48) ÷ (–16)}

= 22 – (1/4) {–5 – 3}

= 22 – (1/4) × –8

= 22 – 1 × –2

= 22 + 2

= 24

1/(3+1/(3+1/(3–1/3)))

= 1/(3 + 1/(3 + 1/(8/3)))

= 1/(3 + 1/(3 + 3/8))

= 1/(3 + 8/27)

= 1/(89/27)

= 27/89

$log_{12} 27$ = a

=> log 27/ log 12 = a

=> $log 3^{3}$ / $log (3 * 2^{2})$ =a

=> 3 log 3 / log 3 + 2 log 2 = a

=> (log 3 + 2 log 2)/ 3 log 3 = 1/a

=> (log 3/ 3 log 3) + (2 log 2/ 3 log 3) = 1/3

=> (2 log 2)/ (3 log 3) = 1/a – 1/3 = (3-a)/ 3a

=> log 2/ log 3= (3-a)/3a

=> log 3 = (2a/3-a)log2

$log_{16} 16 $= log 16/ log 6

= $log 2^{4}/ log (2*3) $

= 4 log 2/ (log 2 + log 3)

= 4(3-a)/ (3+a)