Find the H.C.F of 592 and 252 and express it as a linear combination.
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Using Euclid's division algorithms on 592, 252:
here, 592>252
592=252×2+88
252=88×2+76
88=76×1+12
76=12×6+4
12=4×3+0
Here the remainder is 0.
Thus the H.C.F is 4
here, 592>252
592=252×2+88
252=88×2+76
88=76×1+12
76=12×6+4
12=4×3+0
Here the remainder is 0.
Thus the H.C.F is 4
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