Quotient = 16, Divisor = (25 × 16) = 400
5 × Remainder = Divisor
⇒ Remainder = $$\frac{400}{5}$$ = 80
Divided = (400 × 16) + 80 = 6480

10025 - 25
= (102)25 - 25
= 1050 - 25
$$ = \underbrace {\,1000\,.....00\,}_{50\,{\text{zeros}}} - 25$$     $$ = \underbrace {\,9999\,.....\,9975\,}_{48\,{\text{times}}}$$
∴ Sum of digits
= (48 × 9) + 7 + 5
= 432 + 7 + 5
= 444

From 300 to 399, we note that when '4' comes only one time = 19 such instances.
From 400 to 499, we note that when '4' comes only one time = 80 such instances.
So, total = (19 + 80) = 99 such instances.

Since 774958A96B is divisible by 8, so the number 96B must be divisible by 8.
So, B = 0, as 960 is divisible by 8.
Now, 774958A960 is divisible by 9.
So, (55 + A) must be divisible by 9.
This happens when A = 8
Hence, A = 8 and B = 0

Let a = p - q, b = q - r, c = r - p
∴ a + b + c = p - q + q - r + r - p
⇒ a + b + c = 0
∴ a3 + b3 + c3 = 3abc
⇒ a3 + b3 + c3 = 3(p - q) (q - r) (r - p)

We know (a2 - b2) = (a - b) (a + b)
So, (a2 - b2) ÷ (a + b) = 25
⇒ $$\frac{(a - b) (a + b)}{a + b}$$   = 25
⇒ (a - b) = 25

When n is odd, (xn + an) is always divisible by (x + a).
∴ Each one of (4743 + 4343) and (4747 + 4347) is divisible by (47 + 43)

Let the two - digit number be 10a + b where a > b
According to the question,
a + b = 12 ..... (i)
a - b = 6 ..... (ii)
On adding equation (i) and (ii)
2a = 18
⇒ a = 9
From equation (i),
9 + b = 12
⇒ b = 12 - 9 = 3
∴ Number is 10a + b
= 9 × 10 + 3
= 93
∴ When a < b
Then required number is = 39

Putting n = 1, we get (34n - 43n)
= (34 - 43)
= (81 - 64)
= 17
Which is divisible by 17

Let the two consecutive odd integers be (2m + 1) and (2m + 3)
Then, (2m + 3)2 - (2m + 1)2
= [(2m + 3) + (2m + 1)] [(2m + 3) - (2m + 1)]
= (4m + 4) × 2
= 8m + 8
= 8(m + 1), which is always divisible by 8

z = 13 × 1 + 8 = 21
y = 11 × z + 9 = 11 × 21 + 9 = 240
x = 9 × y + 8 = 9 × 240 + 8 = 2168
Now, when order of divisor is reversed, we have :

∴ Respective remainders are 10, 1 and 6

Unit digit in the given product = Unit digit of (4 × 8 × 7 × 3), which is 2

On dividing 803642 by 11, we get 4 as remainder.
Required number to be added = (11 - 4) = 7

Multiples of 7 between 11 and 200 are 14, 21, 28, 35, 42, ..... , 189, 196
Tm = 196
⇒ 14 + (m - 1) × 7 = 196
⇒ (m - 1) × 7 = 182
⇒ (m - 1) = 26
⇒ m = 27
Multiples of 7 and 3 both, i.e., that of 21 are 21, 42, 63, ....., 189
Tn = 189
⇒ 21 + (n - 1) × 21 = 189
⇒ (n - 1) × 21 = 168
⇒ (n - 1) = 8
⇒ n = 9
∴ Required number of terms = (27 - 9) = 18

Consider the number is 17325
Its unit digit is 5, so it is divisible by 5
Sum of its digits = (5 + 2 + 3 + 7 + 1) = 18, which is divisible by 3
So, the given number is divisible by 3
And since 5 and 3 are co-primes,
So the given number is divisible by (5 × 3), i.e., 15

Given number is 7386038
Sum of its digits = 35, which is not divisible by any of 3 and 9
So, the given number is not divisible by any of 3 and 9
Also, 38 is not divisible by 4
So, the given number is not divisible by 4
Also, (8 + 0 + 8 + 7) - (3 + 6 + 3)
= (23 - 12)
= 11
So, the given number is divisible by 11

10n has (n + 1) digits, Then, 9 will appear n times in (10n - 1).
So, sum of digits in (10n - 1) = 9n
∴ 9n = 4707
⇒ n = $$\frac{4707}{9}$$  = 523

Clearly, (2272 - 875) = 1397, is exactly divisible by N.
Now, 1397 = 11 × 127
∴ The required 3-digit number is 127, the sum of whose digits is 10

Such numbers are 202, 212, 222, 232, 242, 252, 262, 272, 282, 292.
There are 10 such numbers.

The greatest 5-digit number = 99999
On dividing 99999 by 279, we get 117 as remainder
∴ Required number
= (99999 - 117)
= 99882