$x$ ⁿ - 1 will be divisibly by $ x $ + 1 only when $ n $ is even.

(49¹⁵ - 1) = {(7²)¹⁵ - 1} = (7³⁰ - 1), which is divisible by (7 +1), i.e., 8.

Given sum

= 9 +$ \dfrac{3}{4} $+ 7 +$ \dfrac{2}{17} $-$ \left(9 +\dfrac{1}{15} \right) $

= (9 + 7 - 9) +$ \left(\dfrac{3}{4} +\dfrac{2}{17} -\dfrac{1}{15} \right) $

= 7 +$ \dfrac{765 + 120 - 68}{1020} $

= 7 +$ \dfrac{817}{1020} $

Prime numbers less than 50 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Their number is 15

Given Exp. =$ \dfrac{800}{64} $x$ \dfrac{1296}{36} $= 450

On dividing, we get

75) 8485 (113

     75

     ----

      98

      75

      ----

       235

       225

       ----

       10

       ----

Required number = (8485 - 10) // Because 10 < (75 - 10) = 8475.

639 is not divisible by 7

2079 is divisible by each of 3, 7, 9, 11.

On dividing we get,

88) 9217 (104

     88

    ----

     417

     352

     ----

      65

      ----

Therefore, Required number = 9217 + (88 - 65) // Because (88 - 65) < 65.

      = 9217 + 23

      = 9240.

Let 4300731 - $ x $ = 2535618

Then $ x $, = 4300731 - 2535618 = 1765113

Given Exp. =$\dfrac{(a + b)^2 - (a - b)^2}{ab}$=$\dfrac{4ab}{ab}$= 4

Given Exp.= $\left(397\right)$² + $\left(104\right)$² + 2 x 397 x 104

= $\left(397 + 104\right)$²= $\left(501\right)$²

= $\left(500 + 1\right)$²

= $\left(500\right)$ ²+$\left (1\right)$² + $\left(2 \times 500 \times 1\right)$

= 250000 + 1 + 1000= 251001

  35423

+ 7164

+ 41720

  ---------

  84307

- 28213

  ---------

  56094

  ---------

317 x 89 = 317 x $\left(90 -1\right )$

=$ \left(317 \times 90 - 317\right)$

= $\left(28530 - 317\right)$

= 28213

For every natural number $ n $, (x ^{$ n $} - $ a $^{$ n $}) is completely divisible by (x - a).

Let $ x $ = 6$ q $ + 3.

Then, $ x $² = (6q + 3)²

   = 36$ q $² + 36$ q $ + 9

   = 6(6 q ² + 6$ q $ + 1) + 3

Thus, when $ x $² is divided by 6, then remainder = 3.

When $ n $ is even. $\left( x^ n - a ^n\right)$ is completely divisibly by ( x + a )

(17²⁰⁰ - 1²⁰⁰) is completely divisible by (17 + 1), i.e., 18.

$\Rightarrow$    (17²⁰⁰ - 1) is completely divisible by 18.

$\Rightarrow$    On dividing 17²⁰⁰ by 18, we get 1 as remainder.

1400 x $ x $ = 1050   $\Rightarrow$   $ x $ =$ \dfrac{1050}{1400} $=$ \dfrac{3}{4} $

We know that (1² + 2² + 3² + ... + $ n $²) =$ \dfrac{1}{6} n $$\left( n + 1\right)$$\left(2 n + 1\right)$

Putting $ n $ = 10, required sum =$ \left(\dfrac{1}{6} \times 10 \times 11 \times 21\right) $= 385

99 = 11 x 9, where 11 and 9 are co-prime.

By hit and trial, we find that 114345 is divisible by 11 as well as 9. So, it is divisible by 99.

This is an A.P. in which $ a $ = 51, $ l $ = 100 and $ n $ = 50.

$\therefore$Sum =$ \dfrac{n}{2} $$\left( a + l \right)$=$ \dfrac{50}{2} $x (51 + 100)   = 25 x 151   = 3775.

1904 x 1904= (1904)²

= (1900 + 4)²

= (1900)² + (4)² + $\left(2 \times 1900 \times 4\right)$

= 3610000 + 16 + 15200.= 3625216.

Sum of first n natural numbers = $\dfrac{n × (n + 1)}{2}$
Sum of first 40 natural numbers = $\dfrac{40 × 41}{2}$
= 20 × 41
= 820