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11UFind the LCM and HCF of the following integers by the prime factorization meth(1) 12, 15 and 21 (ii) 17, 23, and 29 (1) 8,9 and 25(iv) 72 and 108 (v) 306 and 657 |
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Answer» 12=2×2×315=3×521=3×7HCF=1LCM=2×2×3×5×7=420 right ans LCM i) 12, 15, and 21; 12/2=6/2=3;;15/3=5; 21/3=7; 12)15-12=3)12-12=0, 3x5=15 tuition , ; 17/1=17; 23/1/23; 29/1=29; (iii)8, 9,25, 8/2=4/2=2; 25/5=5; 9/3=3; (iv)72/2=36/2=18/2=9/3=3; 108/2=54/2=27/3=9/3=3; (v)306/2=153/3=51/3=17/1=17; 657/3=219/3=73; |
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