## 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*.

**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.

**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?**

**H.C.F by 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 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 (R_{0})

∴ L = N X Q + R_{0}

Now put the value of L into the expression obtained in step II.

∴ required number will be in the form of (N × Q + R_{0}) K + R

or, (N × Q x K) + (R_{0}K + R)

Clearly N x Q x K is always divisible by N.

Step IV: Now make (R_{0}K + 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).

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G Pugalenthi says

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