Find the H.C.F of 592 and 252 and express it as a linear combination.
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Answered by
3
The H.C.F of 592 and 252 is
592= 253 * 2 + 88,
253= 88 * 2 + 76,
88= 76 * 1 + 12,
76= 12* 6 + 4 ,
12= 4* 3 + 0
therefore HCF of (592,252) is 4
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592= 253 * 2 + 88,
253= 88 * 2 + 76,
88= 76 * 1 + 12,
76= 12* 6 + 4 ,
12= 4* 3 + 0
therefore HCF of (592,252) is 4
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HOPE IT WILL HELP YOU FRIENDS :)
Answered by
5
592 = 252 * 2 + 88
252 = 88 * 2 + 76
88 = 76 * 1 = 12
76 = 12* 6 + 4
12 = 4 * 3 + 0
HCF (592,252) = 4
252 = 88 * 2 + 76
88 = 76 * 1 = 12
76 = 12* 6 + 4
12 = 4 * 3 + 0
HCF (592,252) = 4
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