# GCDs and LCMs

• February 15th 2011, 09:43 PM
amey
GCDs and LCMs
Natural numbers $a,b,c$ are such that :

i) GCD of $a,b$ is $2^33^25^47^1$
ii) GCD of $b,c$ is $2^43^25^5$
iii) GCD of $a,c$ is $2^33^25^4$
iv) Product of abc is $2^1^13^1^05^1^77^2$

Then find the least common multiple (LCM) of $a,b,c$.

• February 16th 2011, 10:42 AM
Opalg
Quote:

Originally Posted by amey
Natural numbers $a,b,c$ are such that :

i) GCD of $a,b$ is $2^33^25^47^1$
ii) GCD of $b,c$ is $2^43^25^5$
iii) GCD of $a,c$ is $2^33^25^4$
iv) Product of abc is $2^1^13^1^05^1^77^2$

Then find the least common multiple (LCM) of $a,b,c$.

It's obvious (I hope) that each of the numbers a, b, c must be of the form $2^?3^?5^?7^?$, 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 $7^1$, and hence the lcm will have a factor of 7.
Now think about the powers of 3. Each pair of numbers has $3^2$ 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 $3^{10}$. So if two of the numbers have $3^2$ then the third one must have $3^6$ (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 $2^?3^65^?7^1$ (where I have left the two ?s for you to find).
Thanks. I got the remaining ones as $2^4$ and $5^8$.