HCF ( 125 , 425 ) = ____
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hcf of 125 &425is 25
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HOLA
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By Euclid's division algorithm
125 < 425
425 = 125 × 3 + 50
125 = 50 × 2 + 25
50 = 25 × 2 + 0
R = 0 , procedure stop
HCF ( 125 , 425 ) = 25
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HOPE U UNDERSTAND ☺☺☺
=====================
By Euclid's division algorithm
125 < 425
425 = 125 × 3 + 50
125 = 50 × 2 + 25
50 = 25 × 2 + 0
R = 0 , procedure stop
HCF ( 125 , 425 ) = 25
==========================
HOPE U UNDERSTAND ☺☺☺
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