Fun of LCM
The lowest common multiple of 5 positive integers is 194040.
Find the minimum possible sum of these 5 numbers.
Bonus:
Solve this problem if we are given 4 or 6 positive integers, instead of 5.
The answer is 82.
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1 solution
Whilst doing this one, I also thought about the problem "What is the lowest sum of 5 integers whose product is 194040?". I reckon this is done by selecting 5 numbers which are as close as possible to 194040^0.2 which is about 11.4. My selection is 14,14,11,10,9 which has sum 58.
For the 4-number scenario I believe the answer would be 9 + 1 1 + 4 0 + 4 9 = 1 0 9 .
For the 6-number scenario we would just add 1 to the 5-number result.
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Can u please prepare a algorithm on that.I am talking about this l.c.m. is consists of 5 primes with raised to some powers.So, for 5 integers we have to select this individual primes for minimum .What about all numbers less than or greater than 5.
I don't understand why the 5 integers must be Coprime? Isn't 3x3 + 5x2 + 2x7 + 2x7 + 11 = 58 A smaller solution to the problem?
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The LCM of 9 , 1 0 , 1 1 and 1 4 is 2 ∗ 3 2 ∗ 5 ∗ 7 ∗ 1 1 = 6 9 3 0 , so the set of 5 numbers you've chosen does not qualify. If it's possible to choose 5 coprime integers then any redundancies in prime factors will be eliminated, ensuring that the desired sum is a minimum while satisfying the LCM condition. It's convenient in this case that 1 9 4 0 4 0 has 5 distinct prime factors, so these 5 primes to their respective powers will provide the smallest sum. In the 4-number scenario, we would still look for 4 coprime integers, but we have to play around with combining 2 of the primes (with powers) to find the least sum. As I mentioned below, combining 2 3 and 5 worked best.
The l.c.m. can be expressed as 1 9 4 0 4 0 = 2 3 . 3 2 . 5 . 7 2 . 1 1
The sum will be minimum if these five integers will be co-prime to each others.If any two of them share a common factor then sum will be increased.So, to attain minimum sum they must be co-prime to each other.
So, the numbers will be for minimum 2 3 , 3 2 , 5 , 7 2 , 1 1 = 8 , 9 , 5 , 4 9 , 1 1 .
Therefore , minimum sum is 8 + 9 + 5 + 4 9 + 1 1 = 8 2