$\frac{x}{2}$ is a perfect square $\Rightarrow \frac{x}{2}=a^2.\frac{x}{3}$ is a perfect cube $\Rightarrow \frac{x}{3}=b^3.\frac{x}{2}\times \frac{x}{3}=a^2\times b^3\Rightarrow x=\sqrt{6a^2{b}^3}=ab\sqrt{6b}.$
For x to be a natural number; 6b has to be a perfect square $\Rightarrow b=6\Rightarrow x=2^3\times 3^4\Rightarrow$ number of divisors $=20$

$(11100{)}_2=0+0+2^2+2^3+2^4=28$

$(1001{)}_x=x^0+x^3=1+x^3$

$1+x^3=28=27+1=3^3+1,x=3.$

We know that Number = Quotient $\times$ Divisor + Remainder

If Divisor (a-2) equals 0, then Number = Remainder. This is the property we are going to use

Make the divisor=0; i.e. put a=2, Then $a^3+3a^2-am+4=m$ where a=2

Thus, 3m= 24 and m=8.

If a number is divisible by 72, it is divisible by 8 and 9. To be divisible by 8, the number formed by the last three digits should be divisible by 8. Therefore, 31Y should be divisible by 8$\Rightarrow$ Y = 2. To be divisible by 9, the sum of the digits of the number should be divisible by 9$\Rightarrow$ X = 3. Therefore, X-Y = 1

Use Chinese remainder theorem,

Let the number be N. N is of the form 9A+2 (i.e. N divided by 9 gives a remainder 2)= 10B+3 (i.e. N divided by 10 gives a remainder 3) =11C+5 (i.e. N divided by 11 gives a remainder of 5)

We will first find the number which is of the form 9A+2=10B+3.--------(1)

Here for B=8, A=9, which are the first integer solutions.

Substituting this in equation (1) we get the value 83, the first number which when divided by 9 gives a remainder 2 and which when divided by 10 gives a remainder 3.

The numbers will fall in an AP with the first digit as 83 and common difference 90 (9*10). The general form of the AP is 83+90k when k=0,1,2.....

Now to include the 3${}^{rd}$ condition(i.e divisor 11 and remainder 5) we can write 83+90k=11C+5 -------(2)

Find the first integer value of K which satisfies the equation such that C is also an integer.

Here for K = 5, C = 48, which are the first integer solutions. Substituting this in equation

(2) we get the value 90 (5) + 83= 533, the first number which when divided by 9 gives a remainder 2, which when divided by 10 gives a remainder 3 and which when divided by 11 gives a remainder 5.

Shortcut:- You may be tempted to mark option (a) after a first glimpse at the options.

Since the question asks for the least number, you should be careful while answering.

You can still back calculate using answer option (a)

option (a) 1523 satisfies the condition. Hence 1523-990 = 533 has to be the first such

number. Answer is option (e)

Also note that, the first number has to be $<$ 990.

option (c)

Numbers which cannot be expressed as difference of squares are all even non-multiples of 4. There are 250 even non-multiples of 4 from 1 to 1000 and 125 such numbers between 1 and 500 (2,6,10......). 125 cannot be expressed hence 375 can be expressed. Hence option c

option (d)

First prime number =2 .To need a zero, we need a 2 and a 5 as $2\times 5=10.$ this occurs only once,

hence there will be only one zero. Answer = option d

Take an example of 11111111. This can be any of the 8 digits.

It would be divisible by 13 and 7 as if we could divide it into equal even groups of 3.

For 19, we need to divide it into even groups of 8. This is not true

This number is divisible by 11, where the rule of divisibility is sum of digits at odd places- sum of digits at even places.

Hence, Answer is none of these.

option(b)

take $N=3$ as $\frac{3}{15}$ gives a remainder 3.

$6N=18\frac{18}{15}$ does not give a remainder 6

$7N=\frac{21}{15}$ remainder =6

$2N=\frac{6}{15}.$ remainder =6

$5N+6=\frac{21}{15}.$ Remainder = 6

option d
$\frac{{10}^1}{13}{\mid }_R=10$
$\frac{{10}^2}{13}{\mid }_R=9$
$\frac{{10}^3}{13}{\mid }_R=(-1)$

Hence we get ${10}^{90}{\mid }_R=1$, we are left with ${10}^1$ which will give a remainder of 10, hence the remainder will be 10 or option d.