Natural numbers are such that :

i) GCD of is

ii) GCD of is

iii) GCD of is

iv) Product of abc is

Then find the least common multiple (LCM) of .

please help me. thank you.

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- Feb 15th 2011, 10:43 PMameyGCDs and LCMs
Natural numbers are such that :

i) GCD of is

ii) GCD of is

iii) GCD of is

iv) Product of abc is

Then find the least common multiple (LCM) of .

please help me. thank you. - Feb 16th 2011, 11:42 AMOpalg
It's obvious (I hope) that each of the numbers a, b, c must be of the form , and that the lcm will be the product of those four primes each raised to the highest of the powers that occur in a, b or c.

You can deal with each of those four primes separately. Take 7 first, because that's the easiest. Each of a, b contains a 7 (to some nonzero power), but the product abc only contains 7 to the power 2. Therefore a, b each contain 7 to the power 1, and c does not contain 7 at all. So the highest power of 7 in the three numbers is , and hence the lcm will have a factor of 7.

Now think about the powers of 3. Each pair of numbers has in its gcd. So each of the three numbers contains 3 to the power at least 2. But if any pair of numbers contained 3 to a power greater than 2 then so would the gcd of those two numbers. Hence two of the numbers a, b, c must contain 3 to the power exactly 2. But the product abc contains . So if two of the numbers have then the third one must have (because 2+2+6=10). Thus the lcm must have 3 to the power 6.

Now do a similar analysis of the powers of 2 and 5, and you will find that the lcm is (where I have left the two ?s for you to find). - Feb 16th 2011, 08:45 PMamey
Thanks. I got the remaining ones as and .