Hcf of 592 and 252 and express it in linear combination
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Step:1 Since 592 > 252 divide 592 by 252
592 = 252 * 2 + 88
Step:2 Since 88 ≠ 0, divide 252 by 88
252 = 88 * 2 + 76
Step:3 Since 76 ≠ 0, divide 88 by 76
88 = 76 * 1 + 12
Step:4 Since 12 ≠ 0, divide 76 by 12
76 = 12 * 6 + 4
Step:5 Since 4 ≠ 0, divide 12 by 4
12 = 4 * 3 + 0
The remainder has now become zero, so we stop here. Since the divisor at this Step is 4, the HCF of 592 and 252 is 4.
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i can't tell step step wise please
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