N = 2 × 4 × 6 × 8 × ..... × 98 × 100
    = 250 (1 × 2 × 3 × ..... × 49 × 50)
    = 250 × 50!
Clearly, the highest power of 2 in N is much higher than that of 5
∴ Number of zeros in N = Highest power of 5 in N
= $$\left[ {\frac{{50}}{5}} \right] + \left[ {\frac{{50}}{{{5^2}}}} \right]$$
= 10 + 2
= 12

Out of four consecutive integers two are even and therefore, their product is even and on adding 1 to it, we get an odd integer. So, n is odd. Some possible values of n are as under :
n = 1 + (1 × 2 × 3 × 4) = (1 + 24) = 25 = 52
n = 1 + (2 × 3 × 4 × 5) = (1 + 120) = 121 = 112
n = 1 + (3 × 4 × 5 × 6) = (1 + 360) = 361 = 192
n = 1 + (4 × 5 × 6 × 7) = (1 + 840) = 841 = 292
Ans so on.....
Hence, n is odd and a perfect square.

Let the missing digit be x.
Then, (80 + x) must be divisible by 4.
Hence, x = 4

461 = 4 × 460 = 4 × (44)15 = 4 × (256)15
Now, (xn - an) is divisible by (x - a) for all values of n.
∴ (25615 - 1) is divisible by (256 - 1) i.e., 255 and hence by 51
⇒ On dividing (256)15 by 51, we get 1 as remainder
⇒ On dividing 460 by 51, we get 1 as remainder
⇒ On dividing 461 by 51, remainder obtained = (4 × 1) = 4

Clearly, 7261 must be replaced by 2961,
which is possible if 6648 is replaced by 6218,
which in turn is possible if 5599 is replaced by 5556.
Thus, the correct number is 555681

Let 6x43 - 46y9 = 1904
Clearly, y = 3 and x = 5
Hence * must be replaced by 5

Let N = 1 × 2 × 3 × 4 × ..... × 1000 = 1000!
Clearly, the highest power of 2 in N very high as compared to that of 5.
So, the number of zeros in N will be equal to the highest power of 5 in N.
∴ Required number of zeros
= $$\left[ {\frac{{1000}}{5}} \right] + \left[ {\frac{{1000}}{{{5^2}}}} \right] + \left[ {\frac{{1000}}{{{5^3}}}} \right]$$      $$ + \left[ {\frac{{1000}}{{{5^4}}}} \right]$$
= 200 + 40 + 8 + 1
= 249

On dividing 11109999 by 1111, we get 9999
So, the required remainder is 1110

(xn + an) is divisible by (x + a) when n is odd
∴ (6767 + 167) is divisible by (67 + 1)
⇒ (6767 + 1) is divisible by 68
⇒ On dividing (6767 + 67) by 68, we get (67 - 1) = 66 as remainder.

Since 13b7 is divisible by 11, we have,
(7 + 3) - (b + 1) = 0
⇒ 9 - b = 0
⇒ b = 9
Putting b = 9, a + 8 = 9 we get a = 1
Hence, (a + b) = (1 + 9) = 10

Number of thumbs = 1, 9, 17, 25, .....
This is an AP in which a = 1 and d = (9 - 1) = 8
∴ Tn = a + (n - 1) d
        = 1 + (n - 1) 8
        = (8n - 7)
∵ 8n - 7 = 1994
⇒ 8n = 2001
⇒ n = 250
T250 = 1 + (250 - 1) 8
        = 1 + 249 × 8
        = 1993
So, 1993 lies on thumb and 1994 on index finger.

Given expression,
$$ = \frac{{\left( {{a^3} + {b^3}} \right)}}{{\left( {{a^2} - ab + {b^2}} \right)}}$$     (where a = 768 and b = 232)
$$\eqalign{ & = \left( {a + b} \right) \cr & = \left( {768 + 232} \right) \cr & = 1000 \cr} $$

Let (46)2 - x2 = 4398 - 3066
Then, (46)2 - x2 = 1332
⇒ x2 = (46)2 - 1332
∴ x2 = (50 - 4)2 - 1332
x2 = (50)2 + 42 - 2 × 50 × 4 - 1332
x2 = 2500 + 16 - 400 - 1332
x2 = 2516 - 1732
x2 = 784
x = 28

34 gives unit digit 1
So, (34)500 gives unit digit 1
And, 33 gives unit digit 7
∴ (13)2003 gives unit digit = (1 × 7) = 7

⇔ p = 5
⇒ (p2 - 1) = (25 - 1)
⇒ (p2 - 1) = 24, which is divisible by 24
⇔ p = 7
⇒ (p2 - 1) = (49 - 1)
⇒ (p2 - 1) = 48, which is divisible by 24
⇔ p = 11
⇒ (p2 - 1) = (121 - 1)
⇒ (p2 - 1) = 120, which is divisible by 24
Hence, (p2 - 1) is always divisible by 24

Perfect squares number between 120 and 300
= 121, 144, 169, 196, 225, 256 and 289
Sum of perfect square between 120 and 300
= 121 + 144 + 169 + 196 + 225 + 256 + 289
= 1400

Given n = $${p_1}^{{x_1}}{p_2}^{{x_2}}{p_3}^{{x_3}}$$
Where p1, p2, p3 are distinct prime factors
Number of prime factors form :
= (x1 × x2 × x3)
= x1 x2 x3
Hence option (B) is correct

Let the required product be n (n + 1) (n + 2)
Then,
n = 1 ⇒ n (n + 1) (n + 2) = (1 × 2 × 3) = 6
n = 2 ⇒ n (n + 1) (n + 2) = (2 × 3 × 4) = 24
n = 3 ⇒ n (n + 1) (n + 2) = (3 × 4 × 5) = 60
So, each such product is divisible by 6

610 × 717 × 1127
= (2 × 3)10 × 717 × 1127
= 210 × 310 × 717 × 1127
Number of prime factors in the given expression
= (10 + 10 + 17 + 27)
= 64

Let the number be x and the quotient be q
Then, x = 195q + 47
= (15 × 13q) + (15 × 3) + 2
= 15 (13q + 3) + 2
So, the given number when divided by 15 gives 2 as remainder.