132 = 4 x 3 x 11
So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
264 ⇒ 11, 3, 4 (/)
396 ⇒ 11, 3, 4 (/)
462 ⇒ 11, 3 (X)
792 ⇒ 11, 3, 4 (/)
968 ⇒ 11, 4 (X)
2178 ⇒ 11, 3 (X)
5184 ⇒ 3, 4 (X)
6336 ⇒ 11, 3, 4 (/)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.

5358 × 51
= 5358 × (50 + 1)
= 5358 × 50 + 5358 × 1
= 267900 + 5358
= 273258

$$\eqalign{ & 23)\,\,\,1056\,\,\,\,(45 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,92 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,136 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,115 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,21 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - \cr & {\text{Required}}\,{\text{number}} \cr & = \left( {23 - 21} \right) \cr & = 2 \cr} $$

Given Exp.
= (397)2 + (104)2 + 2 × 397 × 104
= (397 + 104)2
= (501)2 = (500 + 1)2
= (5002) + (1)2 + (2 × 500 × 1)
= 250000 + 1 + 1000
= 251001

Let the number be 476ab0

476ab0 is divisible by 3
⇒ 4 + 7 + 6 + a + b + 0 is divisible by 3
⇒ 17 + a + b is divisible by 3 - - - - - - - (i)

476ab0 is divisible by 11
[(4 + 6 + b) - (7 + a + 0)] is 0 or divisible by 11
⇒ [3 + (b - a)] is 0 or divisible by 11 - - - - - - - -(ii)

Substitute the values of a and b with the values given in the choices and select the values which satisfies both Equation (i) and Equation (ii).

if a = 6 and b = 2,
17 + a + b = 17 + 6 + 2 = 25 which is not divisible by 3 --- Does not meet equation(i).Hence this is not the answer

if a = 8 and b = 2,
17 + a + b = 17 + 8 + 2 = 27 which is divisible by 3 --- Meet equation(i)
[3 + (b - a)] = [3 + (2 - 8)] = -3 which is neither 0 nor divisible by 11 --- Does not meet equation(ii).Hence this is not the answer

if a = 6 and b = 5,
17 + a + b = 17 + 6 + 5 = 28 which is not divisible by 3 --- Does not meet equation (i) .Hence this is not the answer

if a = 8 and b = 5, 17 + a + b = 17 + 8 + 5 = 30 which is divisible by 3 --- Meet equation 1 [3 + (b - a)] = [3 + (5 - 8)] = 0 ---Meet equation 2 Since these values satisfies both equation 1 and equation 2, this is the answer

$$\eqalign{ & {112 \times {5^4}} \cr & = 112 \times {\left( {\frac{{10}}{2}} \right)^4} \cr & = \frac{{112 \times {{10}^4}}}{{{2^4}}} \cr & = \frac{{1120000}}{{16}} \cr & = 70000 \cr} $$

91 is divisible by 7. So, it is not a prime number.

Let 232 = x.
Then,
(232 + 1) = (x + 1)
Let (x + 1) be completely divisible by the natural number N.
Then,
(296 + 1)
= [(232)3 + 1]
= (x3 + 1)
= (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.

$$\eqalign{ & 1397 \times 1397 \cr & = {\left( {1397} \right)^2} \cr & = {\left( {1400 - 3} \right)^2} \cr & = {\left( {1400} \right)^2} + {\left( 3 \right)^2} - \left( {2 \times 1400 \times 3} \right) \cr & = 1960000 + 9 - 8400 \cr & = 1960009 - 8400 \cr & = 1951609 \cr} $$

Required sum = (2 + 3 + 5 + 7 + 11) = 28
Note: 1 is not a prime number
Definition : A prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 1 and itself.

Let the smaller number be x
Then larger number = (x + 1365)
∴ x + 1365 = 6x + 15
⇒ 5x = 1350
⇒ x = 270
∴ Smaller number = 270

$$\eqalign{ & {\text{Given}}\,{\text{Exp}}{\text{.}} \cr & = \frac{{{{\left( {12} \right)}^3} \times {6^4}}}{{432}} \cr & = \frac{{{{\left( {12} \right)}^3} \times {6^4}}}{{12 \times {6^2}}} \cr & = {\left( {12} \right)^2} \times {6^2} \cr & = {\left( {72} \right)^2} \cr & = 5184 \cr} $$

24 = 3 × 8, where 3 and 8 co-prime
Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8
Similarly, 63810 is not divisible by 8 and 537804 is not divisible by 8
Consider option (D),
Sum of digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3
Also, 736 is divisible by 8
∴ 3125736 is divisible by (3 × 8), i.e., 24

Let the two consecutive even integers be 2n and (2n + 2). Then,
(2n + 2)2 - (2n)2 = (2n + 2 + 2n) (2n + 2 - 2n)
= 2(4n + 2)
= 4(2n + 1), which is divisible by 4

Unit digit in (6374)1793
= Unit digit in (4)1793
= Unit digit in [(42)896 x 4]
= Unit digit in (6 x 4) = 4

Unit digit in (625)317
= Unit digit in (5)317
= 5

Unit digit in (341)491
= Unit digit in (1)491
= 1

Required digit
= Unit digit in (4 x 5 x 1)
= 0

$$\eqalign{ & 935421 \times 625 \cr & = 935421 \times {5^4} \cr & = 935421 \times {\left( {\frac{{10}}{2}} \right)^4} \cr & = \frac{{935421 \times {{10}^4}}}{{{2^4}}} \cr & = \frac{{9354210000}}{{16}} \cr & = 584638125 \cr} $$

Clearly, 97 is a prime number.

72519 × 9999
= 72519 × (10000 - 1)
= 72519 × 10000 - 72519 × 1
= 725190000 - 72519
= 725117481

$$\eqalign{ & x = 13p + 11\,{\text{and}}\,x = 17q + 9 \cr & \therefore 13p + 11 = 17q + 9 \cr & \Rightarrow 17q - 13p = 2 \cr & \Rightarrow q = \frac{{2 + 13p}}{{17}} \cr & {\text{Thr}}\,{\text{least}}\,{\text{value}}\,{\text{of}}\,p\,{\text{for}}\,{\text{which}} \cr & q = \frac{{2 + 13p}}{{17}}\,{\text{is}}\,{\text{a}}\,{\text{whole}}\,{\text{number}}\,{\text{is}}\,p = 26 \cr & \therefore x = \left( {13 \times 26 + 11} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {338 + 11} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 349 \cr} $$

$$\eqalign{ & {\text{Largest}}\,{\text{4 - digit}}\,{\text{number}} = 9999 \cr & 88)\,\,\,\,9999\,\,\,\,(113 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,88 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,119 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,88 \cr & \,\,\,\,\,\,\,\,\,\,\, - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,319 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,264 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,55 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - - - \cr & \text{Required number} \cr & = \left( {9999 - 55} \right) \cr & = 9944 \cr} $$