= 1422 - 1
= (142 - 1) (142 + 1) [a2 - b2 = (a - b) (a + b)]
= 141 × 143
= 141 × 11 × 13

(xyz , xyz) is number
⇒ 100000x + 10000y + 1000z + 100x + 10y + z
⇒ 1001 (100x + 10y + z)
Since 1001 has factors i.e., 7, 11 and 13
So, the answer is 1001

Dividend = (divisor × quotient) + remainder
According to the question
$$\eqalign{ & {\text{Divisor :}} \cr & \Rightarrow {\text{16}} \times {\text{25 = 5}} \times {\text{remainder}} \cr & \Rightarrow {\text{R = }}\frac{1}{5} \times 16 \times 25 \cr & {\text{Dividend :}} \cr & {\text{ = }}\left[ {\left( {{\text{16}} \times {\text{25}}} \right) \times 16} \right] + \frac{1}{5} \times 16 \times 25 \cr & = \left[ {16 \times 25 \times 16} \right] + 80 \cr & = 6480 \cr} $$

$${\text{0}}{\text{.1 = }}\frac{1}{{10}}\left( {{\text{mud}}} \right),\,\,\,\,\frac{5}{8}\left( {{\text{water}}} \right)$$
Let total length of bamboo = 40x ( LCM of 8, 10 )
$$\eqalign{ & {\text{Mud part :}} \cr & {\text{ = }}\frac{1}{{10}} \times 40x = 4x \cr & {\text{Water part :}} \cr & {\text{ = }}\frac{5}{8} \times 40x = 25x \cr & {\text{Remaining part :}} \cr & = (40x - 25x - 4x) \cr & = 11x \cr & 11x{\text{ }} \to {\text{ }}2.75m \cr & {\text{ }}x{\text{ }} \to {\text{ }}0.25m \cr & 40x{\text{ }} \to {\text{ }}10m \cr} $$

Let the total length of rod is 1 unit
According to question,
$$ \Rightarrow 1 - \frac{1}{{10}} - \frac{1}{{20}} - \frac{1}{{30}} - \frac{1}{{40}} - \frac{1}{{50}}$$       $$ - \frac{1}{{60}}$$

$$ \Rightarrow \frac{{600 - 60 - 30 - 20 - 15 - 12 - 10}}{{600}}$$
$$\eqalign{ & \Rightarrow \frac{{453 \leftarrow {\text{ Violet part}}}}{{600 \leftarrow {\text{ Total part}}}} \cr & 453{\text{ part }} \to {\text{ 12}}{\text{.08 mtr}} \cr & {\text{1 part }} \to \,\,\,\,\,\,\,\,\frac{{12.08}}{{453}} \cr & 600{\text{ part }} \to \frac{{12.08}}{{453}} \times 600 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16{\text{mtr}} \cr} $$
So total length of rod is 16 metres.

Let the 3 consecutive number is
x, x + 1, x + 2
According to question,
$$ \Rightarrow {x^2} + {\left( {x + 1} \right)^2} + {\left( {x + 2} \right)^2}$$     = 365
$$ \Rightarrow {x^2} + {x^2} + 1 + 2x + {x^2} + 4 + 4x$$       = 365
$$\eqalign{ & \Rightarrow 3{x^2} + 6x = 360 \cr & \Rightarrow {x^2} + 2x - 120 = 0 \cr & \Rightarrow \left( {x - 10} \right)\left( {x + 12} \right) = 0 \cr & \Rightarrow x = 10 \cr & {\text{Sum of numbers :}} \cr & = 10 + 11 + 12 \cr & = 33 \cr} $$

Let the number is a, b, c
a = 3b       $$\frac{a}{b}$$ = $$\frac{3}{1}$$
c = $$\frac{2}{3}$$a       $$\frac{c}{a}$$ = $$\frac{2}{3}$$

    c         a         b     2 3     3 1 c : a : b = 2 : 3 : 1
So, (2 + 3 + 1) = 252 units
6 units → 252
1 unit → 42
3 units → 42 × 3
So, second number is = 42

Sum of A.P. = $$\frac{n}{2}\left[ {2a + (n - 1)d} \right]$$
nth term = a + (n - 1)d
3rd term :
⇒ a + (3 - 1)d = 19
⇒ a + 2d = 19.....(i)
Sum of 10 term :
$$\eqalign{ & \Rightarrow \frac{{10}}{2}\left[ {2a + 9d} \right] = 390 \cr & \Rightarrow 2a + 9d = 78..... (ii) \cr & \Rightarrow a + 2d = 19 \cr} $$
From equation (i) and (ii)
a = 3
d = 8

$$\eqalign{ & {\text{0}}{\text{.121212}}........ \cr & \Rightarrow 0.\overline {12} = \frac{{12}}{{99}} = \frac{4}{{33}} \cr} $$

Let the integers in a, b, c
a2 + b 2 + c2 = 323.....(i)
a2 + b2 + = 2c.....(ii)
Break the 323 in to their squares
a = 5
b = 3
c = 17
So, a2 + b2 + c2 = 323
and
a2 + b2 = 2c
252 + 32 = 2 × 17
34 = 34
So, Satisfied
a : b : c = 17 × 5 × 3 = 255

$$\eqalign{ & \frac{6}{{\frac{7}{8}}} = \frac{{6 \times 8}}{7} = \frac{{48}}{7} \cr & \frac{{\frac{6}{7}}}{8} = \frac{6}{{7 \times 8}} = \frac{6}{{56}} = \frac{3}{{28}} \cr & {\text{Difference - }} \cr & = \frac{{48}}{7} - \frac{3}{{28}} \cr & = \frac{{192 - 3}}{{28}} \cr & = \frac{{189}}{{28}} \cr & = \frac{{27}}{4} \cr & = 6\frac{3}{4} \cr} $$

Every real number is a rational number

Let numbers are a, a + 1, a + 2
Then,
a + a + 1 + a + 2 = 51
3a + 3 = 51
3a = 48
a = 16
Middle number :
= a + 1
= 16 + 1
= 17

Let the total member be x
x members contribution is Rs. x
And each member also contributes x paise
So, total rupees is
x member Rs. x = Rs. x2
x member Rs. x paise = $$\frac{x.x}{100}$$
Total rupees is
$${x^2} + \frac{{{x^2}}}{{100}} = 2525$$
101x2 = 2525 × 100
101x2 = 101 × 25 × 100
x = 50

$$\eqalign{ & {\text{By option }}\frac{8}{9},{\text{ }} \cr & {\text{Proper fraction = }}\frac{8}{9} \cr & {\text{Its reciprocal = }}\frac{9}{8} \cr & {\text{According to question}} \cr & = \frac{9}{8} - \frac{8}{9} \cr & = \frac{{81 - 64}}{{72}} \cr & = \frac{{17}}{{72}}{\text{ satisfied}} \cr} $$

Smallest number in case of decimal = 0.001

By hit and trial method
= 13851 + 69
= 13920
Which is divisible by 87

$$\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right)\left( {1 - \frac{1}{5}} \right)$$     . . . . .$$\left( {1 - \frac{1}{{99}}} \right)$$ $$\left( {1 - \frac{1}{{100}}} \right)$$
$$\eqalign{ & = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times ..... \times \frac{{98}}{{99}} \times \frac{{99}}{{100}} \cr & = \frac{2}{{100}} \cr & = \frac{1}{{50}} \cr} $$

∵ p = - 0.12
q = - 0.01
r = - 0.015
If all values would be positive
0.12 > 0.015 > 0.01
p > r > q
But these are negative so their order will be
q > r > p

Remaining distance to be run
$$\eqalign{ & = 5 - \frac{5}{4} \cr & = \frac{{20 - 5}}{4} \cr & = \frac{{15}}{4}\,{\text{laps}} \cr} $$