The LCM and HCM of two even numbers are 84 and 2 respectively. Find the passible pair of numbers
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Answered by
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by Euclid division algorithm,
HCF*LCM=product of two numbers
84*2= product of two numbers
168= product of two numbers
HCF*LCM=product of two numbers
84*2= product of two numbers
168= product of two numbers
hardikmishra200:
your answer is wrong man . we have to find possible pairs of number
Answered by
1
factors of 84 is 2×2×3×7
after that multiply 84×2=168
and 168 factors are 2×2×2×3×7
now using hit and trial method
place one side 2×2 and make pair with remaining one
for example:- I have maked two pair
I put one side 2×2 and another side 3×7×2 so one such pair come 4 and 42
and second pair is 2×2×7 and 2×3 so it comes 28 and 6
so like that we can make many more pairs
my pairs of numbers are 4,42 and 28,6
you can verify
see hcf of 4 and 42 is 2 and it's lcm is 84
second one
hcf of 28 and 6 is 2 and it's lcm is 84 so this is also a required pairs
hope it helpful
if it works then please mark me as brainliest
after that multiply 84×2=168
and 168 factors are 2×2×2×3×7
now using hit and trial method
place one side 2×2 and make pair with remaining one
for example:- I have maked two pair
I put one side 2×2 and another side 3×7×2 so one such pair come 4 and 42
and second pair is 2×2×7 and 2×3 so it comes 28 and 6
so like that we can make many more pairs
my pairs of numbers are 4,42 and 28,6
you can verify
see hcf of 4 and 42 is 2 and it's lcm is 84
second one
hcf of 28 and 6 is 2 and it's lcm is 84 so this is also a required pairs
hope it helpful
if it works then please mark me as brainliest
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