Answer: Option A. -> 9
:
A

Ans: (a)

The ${100}^{th}$ and ${1000}^{th}$ position value will be only 1.Divisibility rule for 3 is sum of digits. We already have a sum of 2, we need to make it to 6, 9 or another multiple of 3.

Units digit= a

Tens digit= b

Possible combinations of a+b= 4, 7, 10, 13, 16; where both "a and b" are odd

Thus, the possibility of unit and tens digits are (1, 3), (1, 9), (3, 1), (3, 7), (5, 5), (7, 3),(7, 9), (9, 1), (9, 7).

Answer: Option C. -> $\frac{(\sqrt{13}+1)}{2}$
:
C
Ans :(c)
$x=\sqrt{4+\sqrt{4-x}}$
$x^2=4+\sqrt{4-x};(x^2-4)=\sqrt{4-x}$
Only option c satisfies this condition.
Shortcut : Reverse Gear
$X=\sqrt{4}+a$ small value ( note the minus signs inbetween )
$\Rightarrow$ the answer is something slightly greater than 2 .
Look for an answer option, which is slightly greater than 2
(a) 3 is much greater than 2
(b) $\frac{(\sqrt{13}-1)\sim }{2}1.1$ (this can never be the answer)
(c) $\frac{(\sqrt{13}+1)\sim }{2}2.2$ (this is a possible answer)
(d) $\sqrt{13}=$ something greater than 3. this is also not possible, as the value is too high

Answer: Option D. -> 1
:
D

option (d)

If $P=1!=1$

Then $P+2=3,$ when divided by $2!$ Remainder will be 1.

If $P=1!+2\times 2!=5$

Then, $P+2=7$ when divided by $3!$ Remainder is still 1.

Hence, $P=1!+(2+2!)+(3\times 3!)+.....+(10\times 10!)$

When divided by $11!$ Leaves remainder 1.

Alternative method :
$P=1+2.2!+3.3!+....10.10!$
$=(2-1)1!+(3-1)2!+(4-1)3!+.....+(11-1)10!$
$=2!-1!+3!-2!+...+11!-10!$
$=1+11!$
Hence, the remainder is 1.

Answer: Option D. -> $R>1$
:
D
$R=\frac{{30}^{65}-{29}^{65}}{{30}^{64}+{29}^{64}}$
$\because a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}{b}^2+.....+b^{n-1})$
$\therefore R=\frac{(30-29)[{30}^{64}+({30}^{63}\times 29)+....+{29}^{64}]}{{30}^{64}+{29}^{64}}$
$\because {30}^{64}+{30}^{63}\times 29+...+{29}^{64}>{30}^{64}+{29}^{64}$
$\therefore R>1$
Hence, option 4.
Alternative approach: Unitary Method
$R=\frac{({30}^{65}-{29}^{65})}{({30}^{64}+{29}^{64})}$
If you take this number to be of the form 'n' and 'n+1' instead of 30 and 29, you can solve it using simple numbers like $1\left(n\right)$ and $2(n+1)$
i.e $\frac{(2^3-1^3)}{(2^2+1^2)}=\frac{7}{5}>1$. Only 1 option satisfies this. Option (d)

Answer: Option A. -> 0
:
A

Sol:

$x={16}^3+{17}^3+{18}^3+{19}^3$

$=({16}^3+{19}^3)+({17}^3+{18}^3)$

$=(16+19)({16}^2+16\times 19+{19}^2)+(17+18)({17}^2+17\times 18+{18}^2)$

$=35\times$ (an odd number) $+35\times$ (another odd number)

$=35\times$ (an even number) $=35\times \left(2k\right)..........$ (k is a positive integer)

$x=70k$

Therefore x is divisible by 70. Remainder when x is divided by 70 = 0

Hence, option 1.

Answer: Option B. -> 6
:
B

Option (b):

Let the number be (10y + x)- (10x + y) = 18

$\Rightarrow 9(y-x)=18\Rightarrow y-x=2$

So, the possible pairs of (x, y) are

(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9).

But we want the number other than 13. Thus, there are 6 possible numbers, i.e., 24, 35, 46, 57, 68,79.

So, total number of possible numbers are 6.

Answer: Option C. -> 41
:
C

Option (c)

Sum of 4 consecutive odd numbers should result in a zero in the end, so as to be divisible by 10.

This happens in 7+9+1+3. Using options, we find that four consecutive odd numbers are 37, 39, 41 and 43, as the sum of these 4 numbers is 160, When divided by 10, we get 16 which is a perfect square.

Thus, 41 is one of the odd numbers.

Answer: Option C. -> 9
:
C

Soln:c.
If $p=\frac{1}{3}$ and $q=\frac{2}{3}$, then $p+q=1$ and $pq=\frac{2}{9}$
$a_1+b_1=p+q=1$
$a_3+b_3=p^{2}q+p{q}^2=pq(p+q)=\frac{2}{9}.$
$a_5+b_5=p^3{q}^2+p^2{q}^3=p^2{q}^2(p+q)=\frac{4}{81}.$
So, in general, for odd 'n' we can write
$a_n+b_n={\left(\frac{2}{9}\right)}^{\left(\frac{n-1}{2}\right)}$ ; we need to find least value of n such that ${\left(\frac{2}{9}\right)}^{\left(\frac{n-1}{2}\right)}<0.01$
Using the options given, if $n=7,a_7+b_7={\left(\frac{2}{9}\right)}^3>0.01;$ if $n=9,a_9+b_9={\left(\frac{2}{9}\right)}^4<0.01$

Answer: Option D. -> 6
:
D

Let there be n rows and a students in the first row.

Number of students in the second row = + 3a Number of students in the third row = a+ 6 and so on. aThe number of students in each row forms an arithmetic progression with common difference = 3. The total number of students = The sum of all terms in the arithmetic progression.

$=\frac{n[2a+3(n-1\left)\right]}{2}=630$

Now consider options.

$1.n=3,a=207$

$2.n=4,a=153$

$3.n=5,a=120$

$4.n=6,a=\frac{195}{2}$

$5.n=7,a=8$

Hence the only option not possible is when n=6

Answer: Option D. -> 3
:
D

Soln:d.
Given conditions :
$\left(\frac{1}{m}\right)+\left(\frac{4}{n}\right)=\left(\frac{1}{12}\right)$
M and n are positive integers
N is an odd integer less than 60.
On solving the equation, we get $m=\frac{12n}{(n-48)};$
Since m and n are positive integers, $n>48$, and from the question, $48<n<60$.
Hence possible values on $n=49,51,53,55,57,59.$
For $n=49,51$ and $57$ we get integral values of m. Hence the required no of integral pairs = 3