LCM of 12, 16, 18, 20, 25 = 3600
For remainder 4 = 3600k + 4
If divisible by 7 $$ = \frac{{7 \times 514k + 2k + 4}}{7}$$
$$\frac{{2k + 4}}{7},$$  If k = 5 divisible
x = 3600 × 5 + 4
= 18000 + 4
$$ = \mathop {\mathop {18004}\limits_ \downarrow }\limits_{\boxed8} $$

In this type of questions go through option, we take option B.
$$\frac{{34}}{{13}}$$ remainder is 8.
$$\frac{{34}}{7}$$ remainder is 6.
It satisfy the condition so answer is option B.

$$59535 = 3 \times 5 \times \boxed{3 \times 3} \times \boxed{3 \times 3} \times 7 \times 7$$
⇒ To make a perfect square we should multiply by = 3 × 5 = 15
x = 15 [given]
⇒ Sum of digits of number = 1 + 5 = 6

$$\eqalign{ & 3{x^2} - 4x + 1 = 0 \cr & 3{x^2} - 3x - x + 1 = 0 \cr & \left( {3x - 1} \right)\left( {x - 1} \right) = 0 \cr & 3x - 1 = 0,\,\,x - 1 = 0 \cr & \Rightarrow x = 1,\,\frac{1}{3} \cr & {\text{Put }}x = 1{\text{ in }}p{x^3} - 8{x^2} - qx + 1 \cr & \Rightarrow p - 8 - q + 1 = 0 \cr & \Rightarrow p - q = 7 \cr & {\text{Go through the option C}} \cr & p = \frac{{33}}{4},\,q = \frac{5}{4} \cr & \Rightarrow \frac{{33 - 5}}{4} = 7 \cr & \Rightarrow 7 = 7 \cr} $$

10182 - 1019 × 1017 + 1015 × 1012 - 1016 × 1011
= 10182 - (1018 + 1) × (1018 - 1) + (1016 - 1) × (1011 + 1) - 1016 × 1011
= 10182 - 10182 + 1 + 1016 × 1011 + 1016 - 1011 - 1 - 1016 × 1011
= 5

No explanation is given for this question Let's Discuss on Board

$$\eqalign{ & \frac{{{{77}^{77}} + 77}}{{78}} \cr & = {\left( { - 1} \right)^{77}} + \left( { - 1} \right) \cr & = - 1 - 1 \cr & = - 2 \cr & {\text{Remainder}} = 78 - 2 = 76 \cr} $$

6 < $$\root 3 \of {\text{N}} $$ < 7
On cubing,
216 < N < 343

Number of terms = (342 - 217) + 1 = 125 + 1 = 126

Given: 347XY where X, Y two digit number is completely divisible by 80.
$$\eqalign{ & \frac{{347XY}}{{80}} = \frac{{347XY}}{{8 \times 10}} \cr & \frac{{7 + X + Y}}{8} \cr} $$
∴ Y must be zero
So, go through option A
Put X = 2, then number 3472 is divisible by 8.

$$\eqalign{ & {\text{According to the question,}} \cr & \frac{{3a + 4b}}{2} > 50 \cr & 3a + 4b > 100 \cr & a = 2b\,\,\left( {{\text{Given}}} \right) \cr & 6b + 4b > 100 \cr & 10b > 100 \cr & b > 10 \cr & {\text{If }}b = 10.1 \cr & \therefore a = 2 \times 10.1 \cr & a = 20.2 \cr & \therefore a = 21\left( {{\text{Approx}}} \right) \cr} $$

x is the remainder when 361284 is divided by 5
So, $$\frac{{{3^{61284}}}}{5} = \frac{{{3^{4 \times 15321}}}}{5} = \frac{{{3^4}}}{5} = \frac{{81}}{5}$$
Remainder = 1
x = 1
y is the remainder when 496 is divided by 6
$$\frac{{{4^{96}}}}{6} = \frac{{{4^{4 \times 24}}}}{6} = \frac{{{4^4}}}{6} = \frac{{256}}{6}$$
Remainder = 4
y = 4
Now, (2x - y)
= 2 - 4
= -2

Alternate solution:
As we know,
Unit digit of 361284
⇒ 34
⇒ 1
So we can say if we divide 361284 by 5 then we get remainder 1
x = 1
As we know,
4 ÷ 6, then remainder 4
42 ÷ 6, then remainder 4
43 ÷ 6, then remainder 4
4n ÷ 6, then remainder 4
When n = natural number
So we can say 496 is divided by 6, then the remainder is 4
y = 4
Now, (2x - y)
⇒ 2 × 1 - 4
⇒ 2 - 4
⇒ -2

3600 → 9 × 4 × 25 × 4 = 32 × 24 × 52
Factor: an → (n + 1)
Total number of factors = 3 × 5 × 3 = 45

Let the total member be x.
x member contribute 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
$$\eqalign{ & {x^2} + \frac{{{x^2}}}{{100}} = 2525 \cr & 101{x^2} = 2525 \times 100 \cr & 101{x^2} = 101 \times 25 \times 100 \cr & \boxed{x = 50} \cr} $$

Sum of all prime numbers (x) = 2 + 3 + 5 + 7 + . . . . . . . .
Sum of all odd prime numbers (y) = 3 + 5 + 7 + . . . . . . . .
x - y = 2 Answer

Let the cost of turban be Rs. x
According to the question
$$\frac{{90 + x}}{{12}} = \frac{{65 + x}}{9}$$     (per month pay)
270 + 3x = 260 + 4x
4x - 3x = 10
x = Rs. 10

$$\eqalign{ & \frac{7}{2} + \frac{{11}}{3} + \frac{7}{6} + \frac{{11}}{{15}} + \frac{7}{{12}} + \frac{{11}}{{35}} + \,........\, + \frac{7}{{156}} + \frac{{11}}{{575}} \cr & = \left( {\frac{7}{2} + \frac{7}{6} + \frac{7}{{12}} + \,........\, + \frac{7}{{156}}} \right) + \left( {\frac{{11}}{3} + \frac{{11}}{{15}} + \frac{{11}}{{35}} + \,........\, + \frac{{11}}{{575}}} \right) \cr & = 7\left( {\frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + \,........\, + \frac{1}{{12 \times 13}}} \right) + \frac{{11}}{2}\left( {\frac{2}{{1 \times 3}} + \frac{2}{{3 \times 5}} + \frac{2}{{5 \times 7}} + \,........\, + \frac{1}{{23 \times 25}}} \right) \cr & = 7\left( {1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \,........\, + \frac{1}{{12}} - \frac{1}{{13}}} \right) + \frac{{11}}{2}\left( {1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \,........\, + \frac{1}{{23}} - \frac{1}{{25}}} \right) \cr & = 7\left( {1 - \frac{1}{{13}}} \right) + \frac{{11}}{2}\left( {1 - \frac{1}{{25}}} \right) \cr & = \frac{{7 \times 12}}{{13}} + \frac{{11 \times 24}}{{2 \times 25}} \cr & = \frac{{84}}{{13}} + \frac{{132}}{{25}} \cr & = \frac{{2100 + 1716}}{{325}} \cr & = \frac{{3816}}{{325}} \cr} $$

6 digit number ⇒ abc abc
= abc(1001)
So, number will be divisible by 1001.

$$\eqalign{ & \frac{{\sqrt {29.16} }}{{\sqrt {1.1664} }} + \frac{{\sqrt {0.2916} }}{{\sqrt {116.64} }} + \frac{{\sqrt {0.0036} }}{{\sqrt {0.36} }}\, \cr & = \frac{{5.4}}{{1.08}} + \frac{{0.54}}{{10.8}} + \frac{{0.06}}{{0.6}} \cr} $$

\[\left[ \begin{gathered} \because \sqrt {294} = 54 \hfill \\ \sqrt {11664} = 108 \hfill \\ \sqrt {36} = 6 \hfill \\ \end{gathered} \right.\]

$$\eqalign{ & = \frac{{540}}{{108}} + \frac{{54}}{{1080}} + \frac{6}{{60}} \cr & = 5 + \frac{1}{{20}} + \frac{1}{{10}} \cr & = \frac{{103}}{{20}} \cr} $$

211X
Go through option C
(46)2 = 2116

m12 - 1
= (m6)2 - (1)2
= (m6 + 1)(m6 - 1)
= (m6 + 1){(m3)2 - (1)2}
= (m6 + 1)(m3 - 1)(m3 + 1)
= (m6 + 1)(m3 - 1)(m + 1)(m2 + m + 1)
Hence m + 1 is factor of m12 - 1.
It will completely divide it and give remainder 0.