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Old September 24th, 2009, 09:18 PM
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Default What factors do 4,6,7 have in common

Hey guys, not sure how to state this problem.

I am given the numbers 4,6, and 7. I am then told to find the common numbers of these 3 numbers from 1 to 400? How do I go about it besides listing all the multiples of 4, 6, and 7, and then finding the similar numbers? Thanks for any suggestions!
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Old September 24th, 2009, 10:16 PM
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any common multiples of these three numbers would be a multiple of all three of the numbers. so multiply them
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Old September 24th, 2009, 10:22 PM
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ok, so if I multiply them the first multiple would be 4*6*7 = 168. However, 84 is a common multiple of 4,6, and 7, so it seems this doesn't do the trick. Or am I missing something?
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Old September 25th, 2009, 12:25 AM
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ok, so if I multiply them the first multiple would be 4*6*7 = 168. However, 84 is a common multiple of 4,6, and 7, so it seems this doesn't do the trick. Or am I missing something?
Since 4 \times 6 \times 7 = 2^2 \times 2 \times 3 \times 7 = 2 \times 2^2 \times 3 \times 7, any common multiples of 2^2 \times 3 \times 7 will be a multiple of 4, 6 and 7.
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Old September 25th, 2009, 02:05 AM
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Originally Posted by spearfish View Post
ok, so if I multiply them the first multiple would be 4*6*7 = 168. However, 84 is a common multiple of 4,6, and 7, so it seems this doesn't do the trick. Or am I missing something?
84 is the lowest common multiple of 4,6, & 7. Find all the mutliples of 84 between 84 and 400 = your answer.
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Old September 25th, 2009, 10:51 AM
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Mr_fantastic,

Do you mind explaining why you chose 2^2 * 3 * 7 as the common factors? For example, why did you leave out the other 2? Or why not 2 * 3 * 7 instead of 2^2 * ... ?

I am just trying to figure out a methodical method to solve these types of problems and it seems your answer fits this, but just help me understand it.

Thanks!
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Old September 25th, 2009, 05:05 PM
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Mr_fantastic,

Do you mind explaining why you chose 2^2 * 3 * 7 as the common factors? For example, why did you leave out the other 2? Or why not 2 * 3 * 7 instead of 2^2 * ... ?

I am just trying to figure out a methodical method to solve these types of problems and it seems your answer fits this, but just help me understand it.

Thanks!
OK, here's a better way:

7 is prime. So just get the lowest common multiple of 3 and 4 (which is 12) and then multiply by 7.

Alternatively, 3 is prime so just get the lowest common multiple of 4 and 7 (which is 28) and then multiply by 3.
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Old September 25th, 2009, 08:51 PM
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thanks to all for your time and for your explanations mr_fantastic
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Old September 25th, 2009, 11:44 PM
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adding on to what mr. fantastic had said

when factorized 4*6*7 = 2* 2* 2* 3 *7
he canceled off one of the 2's because in the factorized form u should be able to form each of the original numbers with the least amount of numbers used, in order to get your lowest common multiple

like for 2*2*3*7 you can form a 4 u can also form a 6 and also a 7 therefore this will give u, ur lowest common multiple which is 84
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Old September 25th, 2009, 11:50 PM
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Mr_fantastic,

... why not 2 * 3 * 7 instead of 2^2 * ... ?

Thanks!
Because 4 = 2 \times 2 and so therefore is not a factor of 2 \times 2 \times 3 \times 7.

This method is a methodical way to find the lcm of a set of numbers - break them down into their prime factors and form the number you get by taking the highest indices of those primes that you find in those numbers, and multiply it together.

This is the sort of thing you learn formally in an elementary number theory class at Uni level - it's straightforward to understand but rather trickier to prove.
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Old October 8th, 2009, 04:29 PM
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Thanks again for the explanation guys! It makes perfect sense now
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Old October 9th, 2009, 07:02 AM
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any common multiples of these three numbers would be a multiple of all three of the numbers. so multiply them
This is incorrect. 4 and 6 have a factor of 2 in common so the "least common multiple" of 4 and 6 is 2*2*3= 12. The "least common multiple" of 4, 6, and 7 is 12*7= 84. All common multiples of 4, 6, and 7 are multiples of 84.
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