When n is even, (xn - an) is divisible by both (x - a) as well as (x + a).
Now,
(76n - 66n)
= [(73)2n - (63)2n]
= [(343)2n - (216)2n]
∴ (76n - 66n) is divisible by both (7 - 6) and (7 + 6)
(76n - 66n) is divisible by 13
And, [(343)2n - (216)2n] is divisible by both (343 - 216) and (343 + 216)
⇒ (76n - 66n) is divisible by both 127 and 559

Let the number be N = 2x + 1
N2 = (2x + 1)2
    = 4x2 + 1 + 4x
    = 4x (x + 1) + 1
Clearly, 4x (x + 1) is always divisible by 8 since one of x and (x + 1) is even which when multiplied by 4 is always divisible by 8.
Hence, required remainder = 1

Sum of the digits in 4006020 is 12, which is divisible by 3
Hence, 4006020 is divisible by 3

(248 - 1)
= [(26)8 - 1]
= (64)8 - 1
When n is even , (xn - an) is completely divisible by both (x - a) and (x + a)
∴ (648 - 18) is divisible by both (64 - 1) and (64 + 1)
⇒ (248 - 1) is divisible by both 63 and 65

= (71 × 29 + 27 × 15 + 8 × 4)
= (80 - 9) × 29 + 405 + 32
= (80 × 29) - (9 × 29) + 437
= 2320 - 261 + 437
= 2757 - 261
= 2496

Given expression :
$$ = \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{ab}}$$     (where a = 489, b = 375)
$$\eqalign{ & = \frac{{4ab}}{{ab}} \cr & = 4 \cr} $$

Required number
= (25 - 2)
= (32 - 2)
= 30

Numbers from 501 to 599 which have 6 as digit are 506, 516, 526, 536, 546, 556, 560, 561, 562, 563, 564, 565, 566, 567. 568, 569, 576, 586 and 596, i.e., 6 occurs 20 times.
Number of times 6 occurs from 600 to 699 = 100 + 20 = 120
∴ Total number of times 6 occurs = 20 + 120 = 140

Given x = a (b - c), y = b (c - a), z = c (a - b)
x = a (b - c)
⇒ $$\frac{x}{a}$$ = b - c ..... (i)
Similarly, y = b (c - a)
⇒ $$\frac{y}{b}$$ = c - a ..... (ii)
And similarly z = c (a - b)
⇒ $$\frac{z}{c}$$ = c - a ..... (iii)
Adding (i), (ii) and (iii) we get
∵ $$\frac{x}{a}$$ + $$\frac{y}{b}$$ + $$\frac{z}{c}$$ = b - c + c - a + a - b
⇒ $$\frac{x}{a}$$ + $$\frac{y}{b}$$ + $$\frac{z}{c}$$ = 0
$$\eqalign{ & \therefore {\left( {\frac{x}{a}} \right)^3} + {\left( {\frac{y}{b}} \right)^3} + {\left( {\frac{z}{c}} \right)^3} \cr & = 3 \times \frac{x}{a} \times \frac{y}{b} \times \frac{z}{c} \cr & = \frac{{3xyz}}{{abc}} \cr} $$
[If a + b + c = 0, a3 + b3 + c3 = 3abc]

Sum of digits in the two numbers = 19 + 15 = 34
So, the product will have 33 or 34 digits
Since 36 × 34 = 1224 (i.e., product has 2 + 2 = 4 digits)
So, the number of digits in x is 34

= 394 × 113
= 394 × (100 + 10 + 3)
= (394 × 100) + (394 × 10) + (394 × 3)
= 39400 + 3940 + 1182
= 44522

Let the original number be x.
Then,
$$\frac{(x + 7) × 5}{9}$$   - 3 = 12
⇒ $$\frac{5x + 35}{9}$$  = 15
⇒ 5x + 35 = 135
⇒ 5x = 100
⇒ x = 20

786 × 964
= (800 − 14) × 964
= (800 × 964) − (14 × 964)
= (771200 − 13496)
= 757704

8899 - 6644 - 3322 = x - 1122
⇒ 2255 - 3322 + 1122 = x
⇒ x = 3377 - 3322
⇒ x = 55

= 6 × 66 × 666
= 6 × (6 × 11) × (6 × 111)
= (6 × 6 × 6) × (11 × 111)
= (216 × 1221)
= (1221 × 216)
= 1221 × (200 + 10 + 6)
= (1221 × 200) + (1221 × 10) + (1221 × 6)
= 244200 + 12210 + 7326
= 263736

= 12345679 × 72
= 12345679 × (70 + 2)
= (12345679 × 70) + (12345679 × 2)
= 864197530 + 24691358)
= 888888888

$$\eqalign{ & = \left( {\frac{{16{n^2} + 7n + 6}}{n}} \right) \cr & = \left( {\frac{{16{n^2}}}{n} + \frac{{7n}}{n} + \frac{6}{n}} \right) \cr & = \left( {16n + 7 + \frac{6}{n}} \right) \cr} $$
For $$\left( {16n + 7 + \frac{6}{n}} \right)$$   to be an integer, we may have n = 1 or n = 2 or n = 3 or n = 6
Hence, 4 value of n will give the desired result.

Each of the numbers 302, 303, 304, 305, 306, 308, 309, 310, 312, 314, 315, 316, 318 and 319 is clearly a composite number.
Out of 307, 311, 313 and 317 clearly every one is prime.
Hence, there are 4 prime numbers between 301 and 320.

2361*48 will be divisible by 9 if the sum of the digits of the given number is divisible by 9
2 + 3 + 6 + 1 + * + 4 + 8 i.e., (24 + *) is divisible by 9
Clearly, * = 3 because 27 is divisible by 9

Given expression,
$$ = \frac{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}}{{\left( {{a^2} + {b^2}} \right)}}$$     (where a = 963 and b = 476)
$$\eqalign{ & = \frac{{2\left( {{a^2} + {b^2}} \right)}}{{\left( {{a^2} + {b^2}} \right)}} \cr & = 2 \cr} $$