Problems on H.C.F and L.C.M - General Questions (3)

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.

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, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

 Number to be added = (60 - 37) = 23.

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 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.

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.

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 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.