Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

     = H.C.F. of 48, 92 and 140 = 4.

Clearly, the numbers are (23 x 13) and (23 x 14).

 Larger number = (23 x 14) = 322.

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

 Required number (9999 - 399) = 9600.

Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

 ab = 3.

Now, co-primes with product 3 are (1, 3).

So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).

 Greater number = 111.

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

 The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

Given numbers are 1.08, 0.36 and 0.90.   H.C.F. of 108, 36 and 90 is 18,

 H.C.F. of given numbers = 0.18.

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

 ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

 Required number = (90 x 4) + 4   = 364. 

 2 | 24  -  36  - 40
 --------------------
 2 | 12  -  18  - 20
 --------------------
 2 |  6  -   9  - 10
 -------------------
 3 |  3  -   9  -  5
 -------------------
   |  1  -   3  -  5
   
L.C.M.  = 2 x 2 x 2 x 3 x 3 x 5 = 360.   

L.C.M. of 5, 6, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

 Number to be added = (60 - 37) = 23.

 128352) 238368 ( 1
         128352
         ---------------
         110016 ) 128352 ( 1
                  110016
                 ------------------  
                   18336 ) 110016 ( 6       
                           110016
                           -------
                                x
                           -------
 So, H.C.F. of 128352 and 238368 = 18336.
 
             128352     128352 � 18336    7
 Therefore,  ------  =  -------------- =  --
             238368     238368 � 18336    13          

L.C.M. of 5, 6, 7, 8 = 840.

 Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 Required number = (840 x 2 + 3) = 1683.

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

 L.C.M. of 12, 18, 21 30                 2 | 12  -  18  -  21  -  30
                                         ----------------------------
   = 2 x 3 x 2 x 3 x 7 x 5 = 1260.       3 |  6  -   9  -  21  -  15
                                         ----------------------------
   Required number = (1260 � 2)            |  2  -   3  -   7  -   5

                   = 630.

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

   = 1008 + 7

   = 1015

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

First number =		551		= 19;    Third number =		1073		= 37.
		29				29

 Required sum = (19 + 29 + 37) = 85.

36 = 2² x 3²

84 = 2² x 3 x 7

 H.C.F. = 2² x 3 = 12.

L.C.M. of 8, 16, 40 and 80 = 80.

Required number = (L.C.M. of 12, 15, 20, 54) + 8

   = 540 + 8

   = 548.

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

  = H.C.F. of 1651 and 2032 = 127.

99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

Let the numbers be 2x and 3x.

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

 The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

The required sum =	1	+	1	=	a + b	=	55	=	11
	a		b		ab		600		120