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Re: What is the smallest integer that is divisible by 4, 12, and 15 at the
[#permalink]
16 Dec 2021, 06:37
1
You can easily test for Least Common Multiple (the smallest integer divisible) with a simple process.
First, imagine multiplying all the numbers together: you'd get a big number (OK, it's 720). The reason that there exist smaller numbers than this one that actually do divide is this: there exist redundant factors among 4, 12, and 15. For example, 4 and 12 both divide by 4. 12 and 15 both divide by 3, etc. The numbers are redundant because we don't need to specify that their multiple is divisible, for example, by 4 two times--in fact, that would be saying that it is divisible by 8. That is precisely the outcome we are trying to avoid.
What we want to do is to list out all the factors of each number, individually, and then eliminate all of the redundant factors. If we multiply then, we'll see the smallest number divisible by each of 4, 12, and 15. This would work for any number of numbers, of course.
The process I use is as follows:
1) Prime Factorize:
4: 2^2
12: 2^2 * 3
15: 3 *5
2) Eliminate redundancy: scratch out any number that appears more than once, keeping the highest power of each base.
In this case we get...
4: ___ (nothing)
12: 2^2 * ___ (3 eliminated)
15: 3*5
3) Multiply: the answer is then the remaining numbers multiplied...
2^2 * 3 * 5 = 4 * 15 = 60
Answer B.
And I mean look--you can always just test the numbers. However if you are testing ten numbers instead of three, a process like this will make your life substantially easier. It's worth learning because when you're good at it this is even faster than testing.
Re: What is the smallest integer that is divisible by 4, 12, and 15 at the
[#permalink]
16 Dec 2021, 09:36
Expert Reply
PGTLrowanhand wrote:
You can easily test for Least Common Multiple (the smallest integer divisible) with a simple process.
First, imagine multiplying all the numbers together: you'd get a big number (OK, it's 720). The reason that there exist smaller numbers than this one that actually do divide is this: there exist redundant factors among 4, 12, and 15. For example, 4 and 12 both divide by 4. 12 and 15 both divide by 3, etc. The numbers are redundant because we don't need to specify that their multiple is divisible, for example, by 4 two times--in fact, that would be saying that it is divisible by 8. That is precisely the outcome we are trying to avoid.
What we want to do is to list out all the factors of each number, individually, and then eliminate all of the redundant factors. If we multiply then, we'll see the smallest number divisible by each of 4, 12, and 15. This would work for any number of numbers, of course.
The process I use is as follows:
1) Prime Factorize:
4: 2^2
12: 2^2 * 3
15: 3 *5
2) Eliminate redundancy: scratch out any number that appears more than once, keeping the highest power of each base.
In this case we get...
4: ___ (nothing)
12: 2^2 * ___ (3 eliminated)
15: 3*5
3) Multiply: the answer is then the remaining numbers multiplied...
2^2 * 3 * 5 = 4 * 15 = 60
Answer B.
And I mean look--you can always just test the numbers. However if you are testing ten numbers instead of three, a process like this will make your life substantially easier. It's worth learning because when you're good at it this is even faster than testing.