find the LCM of 9 upon 2, 3 and 21 upon 2
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Step-by-step explanation:
Using prime factorisation method:
(i) 12, 15 and 21
Factor of 12=2×2×3
Factor of 15=3×5
Factor of 21=3×7
HCF (12,15,21)=3
LCM (12,15,21)=2×2×3×5×7=420
(ii) 17, 23 and 29
Factor of 17=1×17
Factor of 23=1×23
Factor of 29=1×29
HCF (17,23,29)=1
LCM (17,23,29)=1×17×23×29=11,339
(iii) 8, 9 and 25
Factor of 8=2×2×2×1
Factor of 9=3×3×1
Factor of 25=5×5×1
HCF (8,9,25)=1
LCM (8,9,25)=2×2×2×3×3×5×5=1,800
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Answer:
Solution: List the prime factors of each. Multiply each factor the greatest number of times it occurs in any of the numbers. 9 has two 3s, and 21 has one 7, so we multiply 3 two times, and 7 once. This gives us 63, the smallest number that can be divided evenly by 3, 9, and 21.
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