$$\eqalign{ & \frac{4}{3},\,\frac{{ - 2}}{9},\,\frac{{ - 7}}{8},\,\frac{5}{{12}} \cr & \frac{4}{3} = 1.33 \cr & \frac{{ - 2}}{9} = - 0.22 \cr & \frac{{ - 7}}{8} = - 0.875 \cr & \frac{5}{{12}} = 0.416 \cr & {\text{So, }}\frac{{ - 7}}{8},\,\frac{{ - 2}}{9},\,\frac{5}{{12}},\,\frac{4}{3} \cr} $$

Let three consecutive odd numbers be x, x + 2 & x + 4
x(x + 2)(x + 4) = 1287
Let take x = 9, x + 2 = 11 & x + 4 = 13 satisfy the equation
Therefore,
Largest number = 13

Let number are a, b
ab = 24 = 2 × 2 × 2 × 3
\[\begin{array}{*{20}{c}} {{\text{Possible}}\left( {a,\,b} \right){\text{pair of}}}&{{\text{Sum}}\left( {a + b} \right)} \\ {\left( {2,\,12} \right)}&{14} \\ {\left( {4,\,6} \right)}&{10} \\ {\left( {8,\,3} \right)}&{11} \\ {\left( {24,\,1} \right)}&{25} \end{array}\]
Then minimum value of the sum = 10

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

$$\eqalign{ & {\text{Sum,}} \cr & {S_n} = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3} \cr & {S_{20}} = \frac{{20 \times 21 \times 22}}{3} \cr & = 140 \times 22 \cr & = 3080 \cr} $$

678p37q is divisible by 75 means it will be divisible by 25 & 3
If 678p37q is divisible by 25 then q = 5
Now check for 3
6 + 7 + 8 + 3 + 7 + q + p
$$ = \frac{{36 + p}}{3}$$
p = 3, 6, 9
Hence, option C is correct

Let the number be 10x + y
Dividend = Divisor × Quotient + Remainder
∴ 10x + y = 6(x + y) + 3
⇒ 10x + y = 6x + 6y + 3
⇒ 10x - 6x + y - 6y = 3
⇒ 4x - 5y = 3 . . . . . . (i)
Again, 10y + x = 4(x + y) + 9
⇒ 10y + x = 4x + 4y + 9
⇒ 6y - 3x = 9
⇒ 2y - x = 3 . . . . . . (ii)
∴ By equation (i) + 4 × (ii),
4x - 5y = 3
8y - 4x = 12
$$\overline {3{\text{y}}\,\,\,\,\,\,\,\,\, = 15} $$
⇒ y = 5
From equation (ii)
2 × 5 - x = 3
⇒ x = 10 - 3
⇒ x = 7
∴ Sum of digits = x + y = 7 + 5 = 12

The greatest 4 digit number = 9999

∴ Number to be added = 307 - 175 = 132

36 - 16 = 20
72 - 52 = 20
80 - 60 = 20
88 - 68 = 20
LCM of 36, 72, 80, 88 = 7920
Least number = 7920 - 20 = 7900
Sum of the digits = 7 + 9 + 0 + 0 = 16

$$\eqalign{ & \frac{c}{a} = \frac{5}{2} \cr & b = \frac{5}{2} - \frac{7}{4} \cr & b = \frac{3}{4} \cr & a + b + c = \frac{{23}}{{12}} \cr & a + c = \frac{{23}}{{12}} - \frac{3}{4} \cr & a + c = \frac{{14}}{{12}} \cr & a + c = \frac{7}{6} \cr & 2x + 5x = \frac{7}{6}\,\,\,\,\,\,\left\{ {\frac{c}{a} = \frac{{5x}}{{2x}}} \right. \cr & 7x = \frac{7}{6} \cr & x = \frac{1}{6} \cr & c - a = 3x = 3 \times \frac{1}{6} = \frac{1}{2} \cr} $$

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

From 400 to 500 = only 466
From 500 to 600 = only 566
From 600 to 700 ⇒ 606, 616, 626, 636, 646, 656, 676, 686, 696 and 660, 661, 662, 663, 664, 665, 667, 668, 669
Total numbers = 20

7, 11, 14, LCM = 154
154k = 154 × 12 = 1848
1848 - 1812 = 36

53 to 97 (Composite number) = 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96
Total number = 35

x = 224, y = 322
⇒ 23x = 23 × 224 = 23 × 16 × 14
⇒ ay = 322a = a × 23 × 14
⇒ If HCF of 23x and ay is divisible by x and y then,
23 × 16 × 14 = a × 23 × 14
a = 16

$$\eqalign{ & {\text{Bucket}} + {\text{full water}} = {\text{17 kg}} \cr & \underline {{\text{Bucket}} + \frac{1}{2}{\text{water}} = 13.5{\text{ kg}}} \to \left( {{\text{Subtract}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{2}{\text{water}} = 3.5\,{\text{kg}} \cr & \therefore {\text{Water}} = 2 \times 3.5 = 7\,{\text{kg}} \cr & \therefore {\text{Weight of empty bucket}} \cr & = 17 - 7 = 10{\text{ kg}} \cr} $$

216 → 23 × 33
= (20 + 21 + 22 + 23) × (30 + 31 + 32 + 33)
Sum of odd factors = 1 + 3 + 9 + 27 = 40

32n + 9n + 5
Put n = 1
⇒ 32 × 1 + 9 × 1 + 5
⇒ 9 + 9 + 5
⇒ 23 ⇒ $$\frac{{23}}{3}$$
⇒ remainder = 2
Note: value of n can be 1, 2, 3, 4, . . . . .

Given:
Two numbers 516 and 525 are divided by 6, the remainders are R1 and R2 respectively.
Concept used:
If a number in the form of (b - 1)n is divided by b
Then,
Remainder = 1 if n is even number
Remainder = (b - 1) if n is an odd number
Calculation:
$$\eqalign{ & {5^{16}} = {\left( {6 - 1} \right)^{16}} \cr & {\text{So, remainder}} = \frac{{{{\left( {6 - 1} \right)}^{16}}}}{6} = 1\,{\text{i}}{\text{.e}}{\text{.,}}\,{{\text{R}}_{\text{1}}} \cr & {\text{Again, }}{5^{25}} = {\left( {6 - 1} \right)^{25}} \cr & {\text{So, remainder}} = \frac{{{{\left( {6 - 1} \right)}^{25}}}}{6} = \left( {6 - 1} \right) = 5\,{\text{i}}{\text{.e}}{\text{.,}}\,{{\text{R}}_{\text{2}}} \cr & {\text{Now,}} \cr & \frac{{{{\text{R}}_1} + {{\text{R}}_2}}}{{{{\text{R}}_2}}} = \frac{{1 + 5}}{5} = \frac{6}{5} \cr & \therefore {\text{Required answer is }}\frac{6}{5} \cr} $$

abcd
Let a = 1
1 2 4 8
÷ 2 ✓
÷ 3 ✓
÷ 13 ✓