Required number = 2047

In 5n we have 5 as unit digit and in 6m we have 6 as unit digit.
∴ 5 + 6 = 11(i.e unit digit always be 1)

Required number = 30005

= 6 × 3 (3 - 1)
= 6 × 3(2)
= 6 × 6
= 36

(325 - 325) = 0.
Which is divisible by 7
So, the given number is divisible by 7
(5 + 3 + 2) - (2 + 5 + 3) = 0
So, the given number is divisible by 11
And,
$$\frac{{325325}}{{13}} = 25025$$
So, 325325 is divisible by 13

If a and b are two rational numbers, then $$\frac{a + b}{2}$$   is a rational number lying between a and b.

We have : (212n - 64n)
= (212n - 24n × 34n)
= 24n (28n - 34n)
Putting n = 1, we get the number 24 (28 - 34)
= 16 (256 - 81)
= (16 × 175)
= 2800
Hence, the number formed by last two digits is 00

$$\eqalign{ & = \left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{7}} \right) + \frac{1}{n} \cr & = \frac{{\left( {21 + 14 + 6} \right)}}{{42}} + \frac{1}{n} \cr & = \left( {\frac{1}{{42}} + \frac{1}{n}} \right) \cr} $$
This sum is a natural number when n = 42
So, each one of the statements that 2 divides ; 3 divides n and 7 divides n is true.
Hence, n > 84 is false

Given expression
$$\eqalign{ & = \frac{{{{\left( {256} \right)}^2} - {{\left( {144} \right)}^2}}}{{112}} \cr & = \frac{{\left( {256 + 144} \right)\left( {256 - 144} \right)}}{{112}} \cr & = \frac{{\left( {400 \times 112} \right)}}{{112}} \cr & = 400 \cr} $$

Prime numbers between 0 and 50 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47
Their number is 15

Let ,   $$\frac{1x5y4}{148}$$ = 78
Then,
10000 + 1000x + 500 + 10y + 4 = 148 × 78
⇒ 10000 + 1000x + 500 + 10y + 4 = 11544
⇒ 10000 + 1000x + 500 + 10y + 4 = 10000 + 1000 + 500 + 40 + 4
⇒ 1000x = 1000
⇒ x = 1

We keep on dividing 33333... by 7 till we get 0 as remainder.

∴ Required number = 47619

⇒ 5P9 - 7Q2 + 9R6 = 823
⇒ (500 + 10P + 9) - (700 + 10Q + 2) + (900 + 10R + 6) = 823
⇒ (500 + 900 - 700) + 10 (P + R - Q) + (9 + 6 - 2) = 823
⇒ 700 + 10 (P + R - Q) = 810
⇒ 700 + 10 (P + R - Q) = 700 + 110
⇒ 10 (P + R - Q) = 110
⇒ P + R - Q = 11
⇒ Q = (P + R - 11)
To get maximum value of Q we take P = 9 and R = 9
This gives Q = (9 + 9 - 11) = 7
Hence, the maximum value of Q is 7

Let the thousand's, hundred's, ten's, and one's digits be x, y, z, w respectively.
Then, the number is
1000x + 100y + 10z + w
= 103x + 102y + 10z + w

Clearly, highest power of 2 is much higher as compared to that of 5 in 60!,
So, Required number of zeros
= Highest power of 5
= $$ \left[ {\frac{{60}}{5}} \right] + \left[ {\frac{{60}}{{{5^2}}}} \right]$$
= 12 + 2
= 14

Let x × 48 = 173 × 240
Then,
x = $$\frac{173 × 240}{48}$$
x = 173 × 5
x = 865

ab = 0
⇒ a = 0 or b = 0 or both are zero

Given ,
a + b + c = 6
ab + bc + ca = 10
∴ (a + b + c)2 = 36
⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 36
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 36
⇒ a2 + b2 + c2 + 2 × 10 = 36
⇒ a2 + b2 + c2 = 16
As we know :
$$ \Rightarrow \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}$$       $$ = \left( {a + b + c} \right)$$
$$\eqalign{ & \Rightarrow \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{16 - \left( {ab + bc + ca} \right)}} = 6 \cr & \Rightarrow \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{16 - 10}} = 6 \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 6 \times 6 \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 36 \cr} $$

The given series is such that the sum of first hundred terms is zero, and 101st term is 2. So, the sum of 101 terms is 2.

Clearly, (D) is true.