Clearly, 100 is divisible by 2. So, 100 is not prime.
(101) < (11)2 and prime numbers less than 11 are 2, 3, 5, 7
Clearly, 101 is not divisible by any of 2, 3, 5 and 7
Hence, 101 is the smallest 3-digit prime number.

m3- 3m2 + 3m + 3n + 3n2 + n3
= - 64 - 48 - 12 - 6 + 12 - 8
= -126

Let $$\frac{46351 - 36418 - 4505}{x}$$     = 1357
Then,
x = $$\frac{46351 - (36418 - 4505)}{1357}$$
x = $$\frac{(46351 - 40923)}{1357}$$
x = $$\frac{5428}{1357}$$
x = 4

Let the unit's digit be x and ten's digit be y
Then, the number is 3422213yx
Also, 99 = (11 × 9), where 11 and 9 are co-primes
Since the given number is divisible by 9, it follows that (3 + 4 + 2 + 2 + 2 + 1 + 3 + y + x) = (17 + y + x) must be divisible by 9
So, y + x = 1 or y + x = 10
Again, the given number is divisible by 11
So, (x + 3 + 2 + 2 + 3) - (y + 1 + 2 + 4) = x - y + 3 is either 0 or 11
∴ (x - y + 3 = 0 or x - y + 3 = 11)
⇒ (y - x = 3 or x - y = 8)
Now, (y + x = 1 and y - x = 3)
⇒ y = 2 and x = - 1
(y + x = 1 and x - y = 8)
⇒ x = $$\frac{9}{2}$$
(y + x = 10 and y - x = 3)
⇒ y = $$\frac{13}{2}$$
(y + x = 10 and x - y = 8)
⇒ x = 9 and y = 1
Thus, x = 9, y = 1
So, required number is 342221319

p < 1
⇒ $$\frac{1}{p}$$ > 1
⇒ $$\frac{2}{p}$$ > 2
⇒ $$\frac{2}{p}$$ - p > 2 - p > 0 [∵ p < 1]
Hence, $$\left( {\frac{2}{p} - p} \right)$$ is a positive number

x + y > p and p > z
⇒ x + y > z

We have 24 = 3 × 8, where 3 and 8 are co-primes
Clearly,
718 is not divisible by 8. So, 35718 is not divisible by 8
810 is not divisible by 8. So, 63810 is not divisible by 8
804 is not divisible by 8. So, 537804 is not divisible by 8
736 is divisible by 8. So, 3125736 is divisible by 8
Also, sum of its digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3
So, 3125736 is divisible by 3 also.
Hence, it is divisible by 24.

a + b + c = 0
⇒ (a + b) = - c
(b + c) = - a
and (c + a) = - b
⇒ (a + b) (b + c) (c + a)
= (- c) × (- a) × (- b)
= - (abc)

Given expression :
= (1234 + 2345 + 4567) - 3456
= (8146 - 3456)
= 4690

= (22 +42 +62 + ..... + 402)
= (1 × 2)2 +(2 × 2)2 +(2 × 3)2 + ..... + (2 × 20)2
= 22 × (12 + 22 + 32 + ..... + 202)
= (4 × 2870)
= 11480

On dividing 1111 by 99, the quotient is 11 and the remainder is 22.
Hence, the required number is 11

(a) Let x = 0 and y = $$\sqrt 2 $$
Then, x is rational and y is irrational.
∴ x + y = 0 + $$\sqrt 2 $$  = $$\sqrt 2 $$ , which is irrational
Thus, x + y is not rational
(b) Let x = 0 and y = $$\sqrt 2 $$
Then, x is rational and y is irrational
∴ xy = 0 × $$\sqrt 2 $$  = 0, which is rational
Hence, xy is not irrational
(c) As shown in (B), xy is not necessarily irrational
(d) x + y is necessary irrational. But xy can be either rational or irrational.
Hence, (D) is true.

(xn + an) is divisible by (x + a) when n is odd
∴ (2525 + 125) is divisible by (25 + 1)
⇒ (2525 + 1) is divisible by 26
⇒ On dividing 2525 by 26, we get (26 - 1) = 25 as remainder

8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 +             8 9 8 7 6 4 8

As proved in the above question,
Divisor = 4375 + 2986 - 2361
             = 5000

The number of prime numbers from 100 to 200 is 21.

101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199.

Unit digit in [(251)98 + (21)29 - (106)100 + (705)35 - 164 + 259]
= unit digit in (1 + 1 - 6 + 5 - 6 + 9)
= 4

We know that (xn - an ) is always divisible by (x - a) for all values of n.

$$\frac{(a - b)}{3.5}$$ = $$\frac{4}{7}$$
⇒ (a - b) = $$\frac{4}{7}$$ × $$\frac{7}{2}$$ = 2
⇒ b < a

Let the unit's place be x and the thousand's place be y.
Then, 357y25x is divisible by 5 only when x = 0 or x = 5
Also, this number is divisible by 3 only when sum of its digits is divisible by 3
So, (22 + x + y) must be divisible by 3
∴ x + y = 2
Taking x = 0, we get y = 2
So, the unit place = 0 and thousand's place = 2