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LCM and HCF Formulas, Shortcuts and Tricks with Examples

March 28, 2019 by Rajashekhar 7 Comments

LCM and HCF Formulas, Shortcuts and Tricks with Examples

LCM and HCF Short Tricks with Examples

Factors and Multiples:

  • If number p divided another number q exactly, we say that p is a factor of q.
  • In this case, q is called a multiple of p.

factors and multiples

Quantitative Aptitude

Common Multiple :
A number which is exactly divisible by all the given numbers is “Common multiple”.

Least Common Multiple (LCM) :
The least number which is exactly divisible by all the given numbers is LCM.
Least Common Multiple (LCM)
Common Factor:
A number which divides all the given numbers exactly is “Common factor”.
Highest Common Factor (HCF):
The greatest number that divides all the given numbers exactly is “HCF”.

H.C.F by Method of Prime Factors:

(1) H.C.F or G.C.F of 18 and 24?
hcf of 18 and 24 by prime factorisation method

H.C.F by Method of Division:

(1) H.C.F of 30 and 42?

HCF of 30 and 42
H.C.F of 30 and 42 = 6

Problems on LCM and HCF

H.C.F and L.C.M of Fractions:
lcm and hcf of fractions
Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.

L.C.M and HCF Important Formulas

  • Product of two numbers (First number x Second Number) = H.C.F. X L.C.M.
  • H.C.F. of a given number always divides its L.C.M.
  • To find the greatest number that will exactly divide x, y and z. Required number = HCF of x, y and z
  • To find the Largest number that will divide x, y and z leaving remainders a, b and c respectively. Required number = HCF of (x -a), (y- b) and (z – c)
  • To find the least number which is exactly divisible by x, y and z. Required number = LCM of x, y and z
  • To find the least number which when divided by x, y and z leaves the remainders a, b and c respectively. It is always observed that, (x – a) = (y – b) = (z- c) = K (say).  Required number = (LCM of x, y and z) – K.
  • To find the least number which, when divided by x, y and z leaves the same remainder r in each case. Required number = (LCM of x, y and z) + r
  • To find the greatest number that will divide x, y and z leaving the same remainder ‘r’ in each case. Required number = HCF of (x -r), (y- r) and (z- r)
  • Largest number which divides x, y, z to leave same remainder = H.C.F. of (y-x), (z-y), (z-x).
  • HCF of two prime numbers is always 1.
  • To find the n-digit greatest number which, when divided by x, y and z,
    (i) leaves no remainder (ie exactly divisible)
    Following step wise methods are adopted.
    Step I: LCM of x, y and z = L
    Step II: L) n-digit greatest number (
    Remainder (R)
    Step III: Required number = n – digit smallest number + (L – R)
    (ii) Leaves remainder K in each case
    Following step-wise method is adopted.
    Step I: LCM of x, y and z = L
    Step II: L) n-digit greatest number (
    Remainder (R)
    Step III: Required number = (n-digit greatest number – R) + K
  • To find the n – digit smallest number which, when divided by x, y and z.(i) Leaves no remainder (i.e. exactly divisible)
    Following steps are followed.
    Step I: LCM of x, y and z = L
    Step II:  LCM) n-digit smallest number (
    Remainder (R)
    Step III: The required number = n-digit smallest number + (L – R)
    (ii) leaves remainder K in each case.
    Step I: LCM of x, y and z = L
    Step II:  LCM) n-digit smallest number (
    Remainder (R)
    Step III: Required number = n – digit smallest number + (L – R) + K
  • To find the least number which on being divided by x, y and z leaves in each case a remiander R, but when divided by N leaves no remainder, following step-wise methods are adopted.
    Step I: Find the LCM of x, y and z say (L).
    Step II: Required number will be in the form of (LK + R); where K is a positive integer.
    Step III: N) L (Quotient (Q)
    —–
    Remainder (R0)
    ∴  L = N X Q + R0
    Now put the value of L into the expression obtained in step II.
    ∴ required number will be in the form of (N × Q + R0) K + R
    or, (N × Q x K) + (R0K + R)
    Clearly N x Q x K is always divisible by N.
    Step IV: Now make (R0K + R) divisible by N by putting the least value of K. Say, 1, 2, 3, 4….
    Now put the value of K into the expression (LK + R) which will be the required number.
  • There are n numbers. If the HCF of each pair is x and the LCM of all the n numbers is y, then the product of n numbers is given by  or Product of ‘n’ numbers = (HCF of each pair)n-1 × (LCM of n numbers).

Filed Under: Aptitude Tagged With: HCM and LCM PDF, LCM and HCF, LCM and HCF Examples, LCM and HCF Formulas, LCM and HCF of Two Papers, LCM and HCF Questions, LCM and HCF Shortcuts

Reader Interactions

Comments

  1. Anushka says

    January 26, 2019 at 1:01 pm

    We gain very important knowledge from your example of LCM,HCF

    Reply
  2. G Pugalenthi says

    January 26, 2019 at 1:22 pm

    Good work

    Reply
  3. Bhushan says

    January 26, 2019 at 3:35 pm

    Nice presentation …

    Reply
  4. Nishika says

    January 27, 2019 at 3:49 am

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    😙😙😙😘😘😘😘😘😎😎😎😎😎

    Reply
  5. K. Jagadeesh says

    January 27, 2019 at 12:35 pm

    Really very good

    Reply
  6. Komala says

    January 28, 2019 at 7:46 am

    Thank you so much for sending this email,it’s very very useful

    Reply
  7. Harmanpreet singh says

    January 28, 2019 at 7:53 am

    Nice solution

    Reply

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