Fast Track Techniques to solve the QUESTIONS of HCF & LCM

1.Product of two numbers = HCF of the numbers x LCM                                                         of the numbers

Q. The LCM of 2 numbers is 2079 and their HCF is 27. If the 1st number is 189, then find the 2nd number.
EXPLANATION Here, LCM = 2079, HCF = 27,
1st number = 189, 2nd number = ?
According to the formula,
              Product of two numbers = HCF x LCM
Therefore
     189 × 2nd number = 2079 × 27
     2nd number = (2079 × 27)/189
                             = 297

2.The least number which when divided by x, y and z leaves the remainders a, b and c respectively, is given by
             [LCM of (x,y, z)] - k. 
             where, k = (x - a) =(y - b) =(z - c)

         

                 Divisibility Tests set 2

Q. Find the least number which when divided by 24, 32 and 36 leaves the remainders 19, 27 and 31, respectively.
EXPLANATION
Given that, x = 24, y = 32, z = 36, a = 19, 6 = 27 and c = 31 Then, 24 - 19 = 5,32 - 27 = 5,36 - 31 = 5
According to the formula,
Required number = (LCM of 24, 32 and 36) -5
Now, LCM of 24, 32 and 36 = 2 x 2 x 2 x 3 x 4 x 3 = 288
∴ Required number = 288 - 5 = 283

3.The greatest number which divides the numbers x, y and z, leaving remainders a, b and c respectively is given by
  HCF of (x-a), (y-b).(z-c)


4.If the two numbers are prime to each other (coprimes), then their HCF should be equal to 1 Conversely, if their HCF is equal to 1, the numbers are prime to each other.

5.LCMof numbers is always divisible by the HCF of given numbers i.e., if x is the LCM of a, b, c, d and e and y is the HCF, then x is completely divisible by y

6.If a is the HCF of two parts of b, then b must be divisible by a. This rule has been derived from the proven fact that, if two numbers are divisible by a certain number, then their sum is also divisible by that number.