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Re: If N is the least positive integer that is a multiple of [#permalink]

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20 Oct 2013, 20:09

ricardochavez wrote:

Well for the sake of the forum, and myself, could you post why exactly 3&5 were ignored?

Posted from my mobile device

To find the lowest common factor of a group of numbers, take the prime factorization of those numbers. For each prime number listed, take the most repeated occurrence of this number in any prime factorization. Then multiply them together.

15 = 3*5; 18 = 2*3^2; 3 occurs twice in 18 (3^2), it only appears one in 15. 40 = 2^3*5; 2 occurs thrice in 40, it only appears once in 18, and once in 50 50 = 2*5^2; 5 occurs twice in 50, it only appears once in 15.

So now that we see the highest power of each prime number that appears in these factorials we multiply them all together:

Re: If N is the least positive integer that is a multiple of [#permalink]

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21 Oct 2013, 12:51

ricardochavez wrote:

Well for the sake of the forum, and myself, could you post why exactly 3&5 were ignored?

Posted from my mobile device

Least Common Multiple = the smallest multiple of two or more integers.

The LCM is the lowest product for all of the shared primes.

We ignore the 3 and 5 because we are looking to find the lowest common multiple.

LCM is not to be confused with the Greatest Common Factor- which is the common factor of ALL integers. They are easily mixed up!

Now to the problem:

Step 1. List the primes for each number

15 - 5,3 18 - 3,3,2 40 - 5,2,2,2 50 - 5,5,2

Step 2. Group common multiples - those that occur the most frequent

5,5(all but 18 have at least one 5 and you must be able to solve for each number given the primes. By having two 5's, we can still solve for 50) 3,3(18 and 15 both have 3, so we need two) 2,2,2 - we need at least three 2's to account for 40. Anything else would be redundant.

Step 3:Multiply together (2^3)(5^2)(3^2) = 1800

B is the only option that fits within the perimeters.
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