Relation between HCF and LCM

The relationship between HCF (Highest Common Factor) and LCM (Least Common Multiple) plays a crucial role in various mathematical applications. those relation between HCF and LCM is stated below:

In case of two numbers the product of two numbers is equal to product of their LCM and HCF, i.e.

(HCF of Two Numbers) × (LCM of Two Numbers) = Product of two Numbers

HCF(A,B) × LCM(A,B) = A × B

While HCF finds the largest number that divides all given numbers, LCM finds the smallest multiple shared by them. The formula connecting them, where the product of HCF and LCM equals the product of the numbers, simplifies many calculations.

Let's Understand the concept with an example,

For Example: 10 and 11 are coprime numbers.

• 10 = 1 × 1
• 11 = 1 × 11

LCM of 10 and 11 = 110
HCF of 10 and 11 = 1
Product of 10 and 11 = 10 × 11 = 110
Consider two numbers A and B, then

Therefore, LCM (A , B) × HCF (A , B) = A × B

Proof of relation between HCF and LCM

To prove : Product of HCF and LCM is equal to product of two numbers

Proof : Let there be two numbers a and b,

Now, a and b can be expressed as the product of their prime factors

a = p1^e1​​p2^e2​​⋯pk^ek and b = a = p1^f1​​p2^f2​​⋯pk^fk

​​The HCF of a and b is found by taking the minimum of the exponents of each prime factor that appears in both factorizations. So:

HCF(a, b) = p1^{min⁡(e1,f1)}p2^{min⁡(e2,f2)}⋯pk^{min⁡(ek,fk)}

Similarly, the LCM of a and b is found by taking the maximum of the exponents of each prime factor that appears in either factorization. So:

LCM(a, b) = p1^{max⁡(e1,f1)}p2^{max⁡(e2,f2)}⋯pk^{max⁡(ek,fk)}

By Multiplying HCF and LCM we get,

HCF(a, b) × LCM(a, b) = ( p1^{min⁡(e1,f1)}p2^{min⁡(e2,f2)}⋯pk^{min⁡(ek,fk)} ) × LCM(a, b) = p1^{max⁡(e1,f1)}p2^{max⁡(e2,f2)}⋯pk^{max⁡(ek,fk)}

Using the properties of exponents, p^m × pⁿ = p^(m+n), we get:

HCF(a, b) × LCM(a, b) = ( p1^{(min⁡(e1,f1)+max⁡(e1,f1) )} p2^{(min⁡(e2,f2)+max⁡(e2,f2))} ⋯pk^{(min⁡(ek,fk)+max⁡(ek,fk))} )

For each prime P_i , we know the following: min⁡(ei,fi)+max⁡(ei,fi)=ei+fi

Thus:
HCF(a,b) × LCM(a,b) = p1^e1+f1p2^e2+f2⋯pk^ek+fk

Which represent the product of prime factorization of a and b

Thus, we can clearly say that:
HCF(a,b) × LCM(a,b) = a × b

This proves the desired relation between the HCF and LCM of two numbers.

Read More,

• HCF and LCM – Aptitude Questions

Relation between HCF and LCM- For three Numbers

The relation between HCF and LCM in cases where three numbers are involved is different.

In case of three numbers we can calculate the LCM by multiplying the numbers and their HCF, and dividing them by their respective pairs' HCF.

The same applies when finding the HCF, we can calculate the HCF by multiplying the numbers and their LCM , and dividing them by their respective pairs' LCM.

Consequently,

For LCM of three Numbers x, y, and z.

LCM (x, y, z)= (x × y × z) × (HCF of x, y, z)/ HCF (x, y) × HCF (y, z) × HCF (x, z)

LCM(x, y, z) = \frac{(x×y×z) × HCF (x, y, z)}{ HCF (x, y) × HCF (y, z) × HCF (x, z)}

For HCF of three Numbers x, y, and z.

HCF (x,y ,z) = (x × y × z) × (LCM of x, y, z)/ LCM (x, y) × LCM (y, z) × LCM (x, z)

HCF(x, y, z) = \frac{(x×y×z) × LCM (x, y, z)}{ LCM (x, y) × LCM (y, z) × LCM (x, z)}

Note: For n numbers the product of HCF and LCM is always less than or equal to the product of the Numbers

HCF(a_1,a_2,a_3,a_4.....a_n) × LCM(a_1,a_2,a_3,a_4,....a_n) \leq a_1×a_2×a_3×a_4×.....a_n

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