Least number of 5 digit is 10000
10000 is divided by 123
then the remainder is 37
Least number that divisible by 123
=10000 − 37 + 123
=10086

Let the numerator and denominator x and y
According to question,
$$\eqalign{ & \frac{{x + 1}}{{y + 1}} = \frac{1}{4} \cr & 4x + 4 = y + 1 \cr & 4x - y = - 3.....(i) \cr & \frac{{x + 2}}{{y + 2}} = \frac{1}{3} \cr & 3x + 6 = y + 2 \cr & 3x - y = - 4.....(ii) \cr & {\text{Solve (i) and (ii) }} \cr & x = 1\,\,\,\& \,\,\,y = 7 \cr} $$
Sum of numerator and denominator of fraction :
= x + y
= 1 + 7
= 8

We know,
Smallest five digit number is = 10000
L.C.M. of 12, 18, 21 = 252
10000 is divided by 252
then the remainder is 172
Difference = 252 - 172 = 80
Then number is = 10000 + 80 = 10080

given $${\text{ }}\frac{3}{4}$$ = 0.75     $${\text{ }}\frac{3}{8}$$ = 0.375
(a) $${\text{ }}\frac{7}{9}$$ → 0.77
(b) $${\text{ }}\frac{7}{3}$$ → 2.3
(c) $${\text{ }}\frac{5}{9}$$ → 0.55
(d) $${1\text{ }}\frac{9}{16}$$ = $${\text{ }}\frac{25}{16}$$ → 1.5
$${\text{ }}\frac{5}{9}$$ = 0.55

$$\eqalign{ & \Rightarrow \left( {0.\overline {63} + 0.\overline {37} } \right) \cr & = \frac{{63}}{{99}} + \frac{{37}}{{99}} \cr & = \frac{{63 + 37}}{{99}} \cr & = \frac{{100}}{{99}} \cr} $$

Since 13 is factor of 52. So divide its remainder directly by 13.
$$\eqalign{ & \Rightarrow \frac{{{\text{remainder}}}}{{13}} = \frac{{45}}{{13}} \cr & \Rightarrow {\text{remainder }} = 6 \cr} $$

In this case:
122 (powers of the last digit 2 )

21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
28 = 256
29 = 512
……

As you can see, the pattern repeats every 4th power (2, 4, 8, 6).

We have to find 122173
Hence, remainder when 173 is divided by 4 = 1
(which corresponds to the 1st number of the pattern)
Thus, unit digit of 122173 = 2

$$999\,\frac{1}{7} + 999\frac{2}{7} + 999\frac{3}{7} + $$     $$999\frac{4}{7} + $$   $$999\frac{5}{7} + $$   $$999\frac{6}{7}$$
$$ \Rightarrow \left( {999 \times 6} \right) + $$   $$\left( {\frac{1}{7} + \frac{2}{7} + \frac{3}{7} + \frac{4}{7} + \frac{5}{7} + \frac{6}{7}} \right)$$
$$\eqalign{ & \Rightarrow 5994 + \left( {\frac{{21}}{7}} \right) \cr & \Rightarrow 5994 + 3 \cr & \Rightarrow 5997 \cr} $$

Let the ten's digits be x
Unit digit be 2x - 1
Original number :
= 10x + 2x - 1
= 12x - 1
New number :
= 10(2x - 1) + x
= 20x - 10 + x
= 21x - 10
According to the question,
(21x - 10) - (12x - 1) = (12x - 1) - 20
9x - 9 = 12x - 21
3x = 12
x = 4
Original number :
= 12x - 1
= 12 × 4 - 1
= 47

Let the number be x
According to question
$$\eqalign{ & \left( {3 \times {x^2}} \right){\text{ - }}\left( {4 \times x} \right) = 50 + x.....(i) \cr & \Rightarrow 3{x^2} - 4x = 50 + x \cr & \Rightarrow 3{x^2} - 5x - 50 = 0 \cr & \Rightarrow 3{x^2} - 15x + 10x - 50 = 0 \cr & \Rightarrow 3{x^{}}(x - 5) + 10(x - 5) = 0 \cr & \Rightarrow (x - 5)(3x + 10) = 0 \cr & x = 5\,\,or\,\, - \frac{{10}}{3} \cr} $$
Since the natural number is x = 5

Shortcut method :
$$ \Rightarrow 3{x^2} - 4x = 50 + x.....(i)$$
Now put the value of x from option (b)
$$\eqalign{ & x = 5 \cr & 3 \times {(5)^2} - 4 \times 5 = 50 + 5 \cr & 75 - 20 = 55 \cr & 55 = 55 \cr} $$
LHS = RHS (it satisfies the conditions)
$${\text{so, x = 5}}$$

$$\eqalign{ & = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{7} + \frac{1}{{14}} + \frac{1}{{28}} \cr & = \frac{{28 + 14 + 7 + 4 + 2 + 1}}{{28}} \cr & = \frac{{56}}{{28}} \cr & = 2 \cr} $$

First multiple = 3
Last multiple (less than 50) = 48
Number of multiples of 3
= $$\frac{48 - 3}{3}$$ + 1
= 16
Sum :
= $$\frac{16}{2}$$ [2 × 3 + (16 - 1)3]
= 8 × 51
= 408

5349 + 3957 - 7062 = 2244
This is not divisible by 7.

a = 4011
b = 3989
a × b = 4011 × 3989
         = 15999879

Shortcut method :
a = 4011 ⇒ (4000 + 11)
b = 3989 ⇒ (4000 - 11)
[(a - b) (a + b) = a2 - b2]
According to the question,
a × b = (4000 + 11) (4000 - 11)
        = 40002 -112
        = 16000000 - 121
        = 15999879

Square root
$$ \Rightarrow \sqrt {\frac{{\left( {0.064 - 0.008} \right)\left( {0.16 - 0.04} \right)}}{{\left( {0.016 + 0.08 + 0.04} \right){{\left( {0.4 + 0.2} \right)}^3}}}} $$
$$ \Rightarrow \sqrt {\frac{{\left[ {{{\left( {0.4} \right)}^3} - {{\left( {0.2} \right)}^3}} \right]\left[ {{{\left( {0.4} \right)}^2} - {{\left( {0.2} \right)}^2}} \right]}}{{\left[ {{{\left( {0.4} \right)}^2} + {{\left( {0.2} \right)}^2} + \left( {0.2 \times 0.4} \right)} \right]{{\left[ {\left( {0.4} \right) + 0.2} \right]}^3}}}} $$
$$ \Rightarrow {\text{Let }}0.4{\text{ }} = {\text{ }}a,{\text{ }}0.2{\text{ }} = {\text{ }}b$$
$$ \Rightarrow \sqrt {\frac{{\left( {{a^3} - {b^3}} \right)\left( {{a^2} - {b^2}} \right)}}{{\left( {{a^2} + {b^2} + ab} \right){{\left( {a + b} \right)}^3}}}} $$
$$ \Rightarrow \sqrt {\frac{{\left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\left( {a + b} \right)\left( {a - b} \right)}}{{{{\left( {a + b} \right)}^3}\left( {{a^2} + {b^2} + ab} \right)}}} $$
$$\eqalign{ & \Rightarrow \sqrt {\frac{{{{\left( {a - b} \right)}^2}}}{{{{\left( {a + b} \right)}^2}}}} \cr & \Rightarrow \frac{{a - b}}{{a + b}} \cr & \Rightarrow \frac{{0.4 - 0.2}}{{0.4 + 0.2}} \cr & \Rightarrow \frac{{0.2}}{{0.6}} \cr & \Rightarrow \frac{1}{3} \cr} $$

5 * 7
= 5 + 7 - 5 × 7
= 12 - 35
= - 23

Sum = 5 + 10 + 15 + 20 + .....+ 95 , It is an arithmetic progression
Number of terms (n)
= $$\frac{{{\text{Last term }} - {\text{ First term}}}}{{{\text{Differnce}}}}$$     + 1
⇒ n = $$\frac{95 - 5}{5}$$ + 1 = 19
So, sum = $$\frac{n}{2}$$ [2a + (n - 1)d]
Here, a = 5, n = 19, d = 5
Sum = $$\frac{19}{2}$$ [10 + (19 - 1)5]
        = 950

$$\eqalign{ & \Rightarrow {2^{16}} - 1\,\, \cr & = \left( {{2^8} - {1^8}\,\,} \right)\left( {{2^8} + {1^8}\,} \right) \cr & = \left( {{2^4} - 1\,\,} \right)\left( {{2^4} + 1\,\,} \right)\left( {{2^8} + 1\,\,} \right) \cr & = \left( {16 - 1} \right)\left( {16 + 1} \right)\left( {{2^8} + 1\,\,} \right) \cr & = 15 \times 17\left( {{2^8} + 1\,\,} \right) \cr & \therefore {2^{16}} - 1\,\,{\text{is divisible by 17}} \cr} $$

Unit's digit in 3 x 38 x 537 x 1256
= Unit's digit in 3 x 8 x 7 x 6
= 4 x 2
= 8
Note : Always multiply only unit digit of first number, to second and product's unit digits number with 3rd number. Again product of last's unit digit to fourth and so on.

Man eats G grapes on 1st day
According to question
G + (G + 6) + (G + 12) + (G + 18) + (G + 24) = 100
5G + 60 = 100
5G = 40
G = 8