LCM and HCF Formulas, Shortcuts and Tricks with Examples

Factors and Multiples:

• If quantity p divided every other quantity q precisely, we are saying that p is a component of q.
• In this case, q is referred to as more than one of p.

 

Least Common Multiple :
A quantity that's precisely divisible through all of the given numbers is “Common more than one”.

Least Common Multiple (LCM) :
The least quantity that's precisely divisible through all of the given numbers is LCM.

https://www.lcm-calculator.com/

Common Factor:

A quantity that divides all of the given numbers precisely is the “Common component”.

Highest Common Factor (HCF):

 

The finest quantity that divides all of the given numbers precisely is “HCF”.

H.C.F through Method of Prime Factors:

(1) H.C.F or G.C.F of 18 and 24?

H.C.F through Method of Division:

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

 

H.C.F and L.C.M of Fractions:

 

Co-primes: Two numbers are stated to be co-prime if their H.C.F. is 1.

 

L.C.M and HCF Important Formulas

• Product of numbers (First quantity x Second Number) = H.C.F. X L.C.M.
• H.C.F. of a given quantity constantly divides its L.C.M.
• To locate the finest quantity in order to precisely divide x, y and z. Required quantity = HCF of x, y and z
• To locate the Largest quantity in order to divide x, y and z leaving remainders a, b and c respectively. Required quantity = HCF of (x -a), (y- b) and (z – c)
• To locate the least quantity that's precisely divisible through x, y and z. Required quantity = LCM of x, y and z
• To locate the least quantity which while divided through x, y and z leaves the remainders a, b and c respectively. It is constantly located that, (x – a) = (y – b) = (z- c) = K (say). Required quantity = (LCM of x, y and z) – K.
• To locate the least quantity which, while divided through x, y and z leaves the identical the rest r in every case. Required quantity = (LCM of x, y and z) + r
• To locate the finest quantity in order to divide x, y and z leaving the identical the rest ‘r’ in every case. Required quantity = HCF of (x -r), (y- r) and (z- r)
• The largest quantity which divides x, y, z to go away identical the rest = H.C.F. of (y-x), (z-y), (z-x).
• HCF of high numbers is constantly 1.
• To locate the n-digit finest quantity which, while divided through x, y and z,
  (i) leaves no the rest (ie precisely divisible)
  Following step smart strategies are adopted.
  Step I: LCM of x, y, and z = L
  Step II: L) n-digit finest quantity (
  The remainder (R)
  Step III: Required quantity = n – digit smallest quantity + (L – R)
  (ii) Leaves the rest K in every case
  Following step-smart technique is adopted.
  Step I: LCM of x, y, and z = L
  Step II: L) n-digit finest quantity (
  The remainder (R)
  Step III: Required quantity = (n-digit finest quantity – R) + K
• To locate the n – digit the smallest quantity which, while divided through x, y, and z.(i) Leaves no the rest (i.e. precisely divisible)
  The following steps are followed.
  Step I: LCM of x, y, and z = L
  Step II: LCM) n-digit smallest quantity (
  The remainder (R)
  Step III: The required quantity = n-digit smallest quantity + (L – R)
  (ii) leaves the rest K in every case.
  Step I: LCM of x, y and z = L
  Step II: LCM) n-digit smallest quantity (
  The remainder (R)
  Step III: Required quantity = n – digit smallest quantity + (L – R) + K
• To locate the least quantity which on being divided through x, y and z leave in every case a reminder R, however, while divided through N leaves no the rest, following step-smart strategies are adopted.
  Step I: Find the LCM of x, y and z say (L).
  Step II: Required quantity can be withinside the shape of (LK + R); wherein K is a tremendous integer.
  Step III: N) L (Quotient (Q)
  —–
  The remainder (R0)
  ∴ L = N X Q + R0
  Now placed the fee of L into the expression received in step II.
  ∴ required quantity can be withinside the shape of (N × Q + R0) K + R
  or, (N × Q x K) + (R0K + R)
  Clearly, N x Q x K is constantly divisible through N.
  Step IV: Now make (R0K + R) divisible through N through placing the least fee of K. Say, 1, 2, three, four….
  Now placed the fee of K into the [removed]LK + R) with the intention to be the desired quantity.
• There are n numbers. If the HCF of every pair is x and the LCM of all of the n numbers is y, then the manufactured from n numbers is given through or Product of ‘n’ numbers = (HCF of every pair)n-1 × (LCM of n numbers).

Frequently Asked Questions on Factors, LCM (LCD) or GCF (GCD or HCF)

What is the distinction between LCM and HCF?

LCM stands for Lowest Common Multiple, and HCF stands for Highest Common Factor.

The LCM of integers is the smallest complete quantity that looks in each in their instances tables, this is, the smallest integer that may be a more than one of each numbers.

For example, the LCM of four and six is 12.

The HCF of integers is the biggest complete quantity that divides each number with out leaving the rest.

For example, the HCF of sixteen and 32 is sixteen.

What is HCF with an example?

HCF or Highest Common Factor of or greater numbers is the finest component that divides the numbers. For example, three is the HCF of three and six.

What is LCM with an example?

LCM or Least not unusualplace more than one is the smallest quantity that's divisible through or greater given numbers. For example, LCM of 2 & four is four.

What is the GCF of 24 and 36?

By high factorisation, we know; Factors of 36 = 1, 2, three, four, 6, 9, 12, 18 and 36; Factors of 24 = 1, 2, three, four, 6, 8, 12 and 24; HCF of (24,36) = 12

What is the formulation for HCF and LCM?

Product of Two numbers = (HCF of the 2 numbers) x (LCM of the 2 numbers)

How are we able to locate the LCM and HCF?

We can locate LCM and HCF the usage of high factorization and lengthy department technique

What is Multiple?

The simple definition of more than one is manifold. In math, the means of more than one is the product end result of 1 quantity accelerated through every other quantity.

Ex: 7 x 8 = fifty six.

Here, fifty-six is more than one of the integer 7.

What is Factor?

Multiplying complete numbers offer a product. The numbers that we multiply are the elements of the product.

three × five = 15 therefore, three and five are the elements of 15.

This additionally means:

A component divides various absolutely without leaving any rest.

For example 30 ÷ 6 = five, and there may be no the rest. So we will say that five and six are the elements of 30.