Answer: Option E. -> None of these
:
E
10 factors implies that the number can be of the form
1) $a^9$ ( Least odd number will be $3^9$)
2) $a{b}^4$ (Least odd number will be $3^4\ast 5=405$)
Therefore, as the least number is 405, there are no two digit numbers of the specified form. The answer is 0.
If a number N can be factorized as $N=a^m\ast b^n$(a, b are the prime factors)
Then number of factors= (m+1)(n+1)

Answer: Option B. -> 8
:
B
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.

Answer: Option E. -> none of these
:
E
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.

Answer: Option C. -> 20
:
C
$\frac{x}{2}$ 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 $=20$

Answer: Option D. -> 17
:
D
The first power of 10 for which a division by 13 gives a remainder -1 is x=3. the next power of 10 will be ${10}^9$ . hence it falls in an AP with first term as 3 and last term as 99 with a common difference of 6
number of terms $=\frac{(99-3)}{6}+1=\frac{96}{6}+1=16+1=17$ terms.

Answer: Option A. -> 8
:
A

 Option A.

Given that the combined average is M+F, hence

 $(M+F)=\frac{(2MF+42)}{\left\{\right(M+F)+1\}}$

Or  $(M+F{)}^2+(M+F)=2MF+42$

Or  $M^2+F^2+M+F=42$

Or $M(M+1)+F(F+1)=42,$  since M and F are integers, there is only one possibility that M(M+1) =12 or 30 and F(F+1) =30 or 12. Hence M can be 3 or 5 and F can be 5 or 3. So the sum of M and F is 8. Thus option (a).

Answer: Option D. -> Can't be determined
:
D
Option D :
The value of F = 5 or 3; so, we can’t determine exact value of F. Answer is can’t be determined

Answer: Option D. -> 729
:
D
$5^y-\frac{5^y}{5}=100;5^y(1-\frac{1}{5})=5^2\times 4;5^y\left(\frac{4}{5}\right)=5^2\times 4$
$⟹5^{(y-1)}\times 4=5^2\times 4;y-1=2;y=3.$ Answer is $(3y{)}^y=(9{)}^3=729$.

Answer: Option B. -> 118
:
B

$1^{st}$ Method: - $71\times 47=3337$ and $17\times 47=799$, so $ab+cd=118$ (Option b).

$2^{nd}$ Method: - Odd $\times$ Odd = Odd and Odd + Odd = Even.

From the question we can conclude that all the four numbers are odd numbers. So the required answer has to be even, only option (b) is even.

Answer: Option A. -> 0
:
A

$\frac{\left(11111\right)}{4}+\frac{5^{11111}}{4}=\frac{3}{4}+\frac{(4+1{)}^{11111}}{4}=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}.$ Thus the required remainder is 0.