Note: It can be extended for all integers, except zero i.e., b ≠ 0.
Example 1: Find HCF of 56 and 72.
Solution:
Apply lemma to 56 and 72.
Take a bigger number and locate ‘b’ and ‘r’. 72 = 56 × 1 + 16
Since 16 ≠ 0, consider 56 as the new dividend and 16 as the new divisor. 56 = 16 × 3 + 8
Again, 8 ≠ 0, consider 16 as the new dividend and 8 as the new divisor. 16 = 8 × 2 + 0.
Since the remainder is zero, divisor 8 is HCF.
Example 2: There are two positive numbers such that one exceeds the other by 20 and their LCM and HCF are 504 and 4 respectively. Find the numbers.
Solution Let the numbers be 'x' and 'x + 20'.
We know that:
HCF (p,q) x LCM (p, q) = p x q
x(x + 20) = 4 x 504
x^{2} + 20x = 2016
x^{2} + (56  36)x  2016 = 0
(x + 56)(x  36) = 0
x = 56 & x = 36
x cannot be negative.
So two numbers are: x = 36 & (x + 20) = 56
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