find HCF of i) 25 and 70 ii)120 and 72 iii)1288 and 572 using euclids division lemma
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3. 1288 = 2 × 575 + 138
575 = 4 ×138 + 23
138 = 6 × 23 + 0
Hence the HCF of 1288 and 575 is 23.
1. 50 and 70
70=1x50+20
50=20x2+10
20=10x2+0
HCF=10
2. 120 = 72 x 1 + 48
72 = 48 x 1 + 24
48 = 24 x 2 + 0
therefore HCF of 120 and 72 = 24.
575 = 4 ×138 + 23
138 = 6 × 23 + 0
Hence the HCF of 1288 and 575 is 23.
1. 50 and 70
70=1x50+20
50=20x2+10
20=10x2+0
HCF=10
2. 120 = 72 x 1 + 48
72 = 48 x 1 + 24
48 = 24 x 2 + 0
therefore HCF of 120 and 72 = 24.
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