Given expression :
= 884697 - (773697 +102479)
= 884697 - 876176
= 8521

$$\eqalign{ & \Leftrightarrow {n^{65}} - n \cr & = n\left( {{n^{64}} - 1} \right) \cr & = n\left( {{n^{32}} - 1} \right)\left( {{n^{32}} + 1} \right) \cr & = n\left( {{n^{16}} - 1} \right)\left( {{n^{16}} + 1} \right)\left( {{n^{32}} + 1} \right) \cr} $$
  $$ = n\left( {{n^8} - 1} \right)\left( {{n^8} + 1} \right)\left( {{n^{16}} + 1} \right)$$     $$\left( {{n^{32}} + 1} \right)$$
  $$ = n\left( {{n^4} - 1} \right)\left( {{n^4} + 1} \right)\left( {{n^8} + 1} \right)$$     $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$
  $$ = n\left( {{n^2} - 1} \right)\left( {{n^2} + 1} \right)\left( {{n^4} + 1} \right)$$     $$\left( {{n^8} + 1} \right)$$  $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$
  $$ = \left( {n - 1} \right)n\left( {n + 1} \right)\left( {{n^2} + 1} \right)\left( {{n^4} + 1} \right)$$       $$\left( {{n^8} + 1} \right)$$  $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{64}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$
Clearly, (n - 1), n and (n + 1) are three consecutive numbers and they have to be multiples of 2, 3 and 4 as n is odd.
Thus, the given number is definitely a multiple of 24

$$\eqalign{ & = 1260 \div 14 \div 9 \cr & = \left( {1260 \times \frac{1}{{14}} \times \frac{1}{9}} \right) \cr & = 10 \cr} $$

B > A ⇒ A < B ⇒ (A - B) < 0
Since A and B are both positive integers, we have (A + B) > 0 and AB > 0
If (A = 1 and B = 2), we have AB < (A + B)
If (A = 2 and B = 3), we have (A + B) < AB
Thus, we cannot say which one of A + B and AB has the highest value.

Statement 1. Let x = 4 and y = 15. Then, each one of x and y is a composite number.
But, x + y = 19, which is not composite.
∴ Statement 1 is not true.

Statement 2. We know that 1 is neither prime nor composite.
∴ Statement 2 is not true.
Thus, neither 1 nor 2 is correct.

307 × 225 × 3411
= (2 × 3 × 5)7 × (2 × 11)5 × (2 × 7)11
= 2(7 + 5 + 11) × 37 × 57 × 115 × 1711
= (223 × 37 × 57 × 115 × 1711)
Number of prime factors
= (23 + 7 + 7 + 5 + 11)
= 53

0 < x < 1
⇒ x2 < x < 1 ..... (i)
⇒ $$\frac{1}{{{x^2}}}$$ > $$\frac{1}{x}$$ > 1 > x > x2 [using ...(i)]
Hence, $$\frac{1}{{{x^2}}}$$ is the greatest.

11, 111, 1111, 11111,.....
Let m = 2 ⇒ 4 × 2 + 3 = 11
m = 27 ⇒ 4 × 27 + 3 = 111
Each number can be expressed in the form (4m + 3) where m is a natural number.
Hence, statement 1 is only correct.

Let 74844 ÷ x = 54 × 63
Then, $$\frac{74844}{x}$$ = 54 × 63
⇒ x = $$\frac{74844}{54 × 63}$$
⇒ x = $$\frac{8316}{6 × 63}$$
⇒ x = $$\frac{1386}{63}$$   = $$\frac{198}{9}$$
⇒ x = 22

Let the dividend be (x + 71) and the divisor be y.
Then, [2(x + 71) - 43] is divisible by y
⇒ (2x + 142 - 43) is divisible by y
⇒ (2x + 99) is divisible by y
∴ Divisor = 99

Shortcut method
Divisor = (2 × 71 - 43) = (142 - 43) = 99

Let the required digit be x
Then,
⇒ x + 1x + 2x + x3 + x1 = 21x
⇒ x + 10 + x + 20 + x + 10x + 3 + 10x + 1 = 200 + 10 + x
⇒ 22x = 210 - 34 = 176
⇒ x = 8
Hence, the required digit is 8

Let 414 × x × 7 = 127512
Then,
$$\eqalign{ & x = \frac{{127512}}{{414 \times 7}} \cr & \,\,\,\,\,\, = \frac{{18216}}{{414}} \cr & \,\,\,\,\,\, = \frac{{2024}}{{46}} \cr & \,\,\,\,\,\, = \frac{{1012}}{{23}} \cr & \,\,\,\,\,\, = 44 \cr} $$

Let 10531 + 4813 - 728 = x × 87
Then, (15344 - 728) = 87 × x
x = $$\frac{14616}{87}$$ = 168

In 978626, we have (6 + 6 + 7) - (2 + 8 + 9) = 0
Hence, 978626 is completely divisible by 11

= (217 × 217) + (183 × 183)
= (217)2 + (183)2
= (200 + 17)2 + (200 - 17)2
= (a + b)2 + (a - b)2
= 2(a2 + b2), where a = 200, b = 17
= 2 [(200)2 + (17)2 ]
= 2 [40000 + 289]
= (2 × 40289)
= 80578

(461 + 462 + 463 + 464)
= 461(1 + 4 + 42 + 43)
= 461 × 85,
Which is divisible by 17

Given expression:
= 38649 - (1624 + 4483)
= (38649 - 6107)
= 32542

Product of all odd natural numbers less than 5000
$$\eqalign{ & = 1 \times 3 \times 5 \times 7 \times ..... \times 4999 \cr & = \frac{{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times ... \times 5000}}{{2 \times 4 \times 6 \times 8 \times ... \times 5000}} \cr & = \frac{{5000!}}{{{2^{2500}}\left( {1 \times 2 \times 3 \times 4 \times ... \times 2500} \right)}} \cr & = \frac{{5000!}}{{{2^{2500}} \times 2500!}} \cr} $$

Let P = n (n + 1) (n + 2) (n + 3)
Then, n = 1 gives
P = (1 × 2 × 3 × 4) = 24
Hence, the required number is 24

Let the numbers be a and b
∴ a + b = 14.....(i)
a - b = 10..... (ii)
From equation (i) and (ii)
2a = 24
a = 12
and b = 2
Product = 12 × 2 = 24
⇒ ab = 24