find the lcm and hcf of the following pairs of integers and verify lcm and hcf = product of two numbers.
(i) 26 and 91
(ii)510 and 92
(iii)336 and 54
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find the lcm and hcf of the following pairs of integers and verify lcm and hcf = product of two numbers.
(i) 26 and 91
(ii)510 and 92
(iii)336 and 54
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[tex](i)LCM\ of\ 26\ and\ 91=182\\HCF\ of\ 26\ and\ 91=13\\Now,\\LCM×HCF=182×13=2366\\and,\\Product(26,91)=26×91=2366.\\Therfore,\\LCM×HCF=Product\\Hence\ verified.\\\\(ii)LCM(510,92)=23460\\HCF(510,92)=2\\Now,\\LCM×HCF=23460×2=46920\\and,\\Product(510,92)=46920\\Therefore,\\LCM×HCF=Product\\Hence\ Verified.\\\\(iii)LCM(336,54)=3024\\HCF(336,54)=6\\Now,\\LCM×HCF=3024×6=18144\\and,\\Product(336,54)=18144\\Therefore,\\LCM×HCF=Product\\Hence\ verified.[/tex]
Verified answer
Answer:
1. By prime factorisation
26=2*13
91=7*13
Therefore HCF is 13
and LCM is 2*7*13=182
Now,
Lcm*Hcf=Product of the number
13*182=26*91
2366=2366
Hence,LHS=RHS
2.By prime factorisation,
510=2*3*5*17
92=2*2*23
Therefore, HCF is 2
and LCM is 2*2*3*5*17*23=23460
Now,
Lcm*Hcf=product of the number
2*23460=510*92
46920=46920
Hence, LHS=RHS
3.By prime factorisation,
336=2*2*2*2*3*7
54=2*3*3*3
Therefore,HCF is 2*3=6
and LCM is 2*2*2*2*3*3*3*7=3024
Now,
Lcm*hcf=product of the number
3024*6=336*54
18144=18144
Hence,LHS=RHS
Hence, verified .
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