Question

Find the HCF of 592 & 252 and express it as a linear combination of them.

Answer 1

$\brown{using\:Euclid's\:division\:lemma}$

$592 = 252 * 2 + 88----eqn(1)\\\\252 = 88 * 2 + 76----eqn(2)\\\\88 = 76 * 1 + 12----eqn(3)\\\\76 = 12 * 6 + 4---eqn(4)\\\\12 = 4*3+0$

$\brown{HCF\:is\:4}$

now,

by eqn (4)

4 = 76-12*6

by eqn (3)

4=76 +(76*1-88)*6

4 = 76 + 76 *6 - 88*6

4= 76 (7) +88(-6)

by eqn (2)

4 = (252-88*2)(7)+88(-6)

by eqn (1)

4=(252-(592-252*2)*2)(7)+(592-252*2)(-6)

4=(252-592*2+252*4)(7) + 592(-6)+252(12)

4=252(7)+592(-14)+252(28)+592(-6)+252(12)

4 = 252 (7+28+12) + 592 (-14-6)

4 = 252 ( 47 ) + 592 ( -20 )

HCF is represented as linear combination in two variables

4 = 252 m + 592 n

where, m = 47 and n = - 20.