How To Solve Lcm Problems

How To Solve Lcm Problems-41
Example Find least common multiple (lcm) of 21 and 48 Solution Step 1: The prime factors of 21 and 48 are 21 = 3 × 7 48 = 2 × 2 × 2 × 2 × 3 Step 2: Maximum occurrences of the prime factors are 2(4 times); 3(1 time); 7(1 time) Step 3: Least common multiple of 21 and 48 = 2 × 2 × 2 × 2 × 3 × 7 = 336 A bell rings every 18 seconds, another every 60 seconds. At what time will the bells ring again at the same time?Step 1: A bell rings every 18 seconds, another every 60 seconds Prime factorizations of 18 and 60 are 18 = 2 × 3 × 3 60 = 2 × 2 × 3 × 5 Step 2: LCM is the product of maximum occurrences of each prime factor in the given numbers.

Example Find least common multiple (lcm) of 21 and 48 Solution Step 1: The prime factors of 21 and 48 are 21 = 3 × 7 48 = 2 × 2 × 2 × 2 × 3 Step 2: Maximum occurrences of the prime factors are 2(4 times); 3(1 time); 7(1 time) Step 3: Least common multiple of 21 and 48 = 2 × 2 × 2 × 2 × 3 × 7 = 336 A bell rings every 18 seconds, another every 60 seconds. At what time will the bells ring again at the same time?Step 1: A bell rings every 18 seconds, another every 60 seconds Prime factorizations of 18 and 60 are 18 = 2 × 3 × 3 60 = 2 × 2 × 3 × 5 Step 2: LCM is the product of maximum occurrences of each prime factor in the given numbers.

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Well, this one might pop out at you, because 36 itself is a multiple of 12.

The least common multiple of 18 and 12 is equal to question mark. So there's a couple of ways you can think about-- so let's just write down our numbers that we care about. So there's two ways that we could approach this. We can take the prime factorization of both of these numbers and then construct the smallest number whose prime factorization has all of the ingredients of both of these numbers, and that will be the least common multiple. 18 is 2 times 9, which is the same thing as 2 times 3 times 3, or 18 is 2 times 9. So we could write 18 is equal to 2 times 3 times 3.

So there's a couple of ways we could think about it. So what's the largest number that divides into both of them? We could list all of their normal factors and see what is the greatest common one. So let's just do the prime factorization method. And then 30 pencils, so I'll just do those in green.

Let's think about what the greatest common divisor of both these numbers are.

So another way to say this is LCM, in parentheses, 36 to 12.

And this is literally saying what's the least common multiple of 36 and 12? So the smallest number that is both a multiple of 36 and 12-- because 36 is a multiple of 12-- is actually 36. And they just state this with a different notation. Now, the least common multiple of 18 and 12-- let me write this down-- so the least common multiple of 18 and 12 is going to have to have enough prime factors to cover both of these numbers and no more, because we want the least common multiple or the smallest common multiple. Well, it needs to have at least 1, 2, a 3 and a 3 in order to be divisible by 18. Luis's teacher also assigns three projects per year. And we could keep going on and on looking at all the multiples of 30. Will William's teacher, after the first test, they're going to get to 24 questions.Even though the two classes have to take a different number of exams, their teachers have told them that both classes-- let me underline-- both classes will get the same total number of exam questions each year. So this is probably a hint of what we're thinking about. We want the minimum multiples or the least multiple. Then they're going to get to 48 after the second test. They're going to get to 96 after the fourth test. So another way to come up with the least common multiple, if we didn't even do this exercise up here, says, look, the number has to be divisible by both 30 and 24. And say, well in order to be divisible by 24, its prime factorization is going to need 3 twos and a 3. And we already have 1 two, so we just need 2 more twos. So this makes it-- let me scroll up a little bit-- this right over here makes it divisible by 24. She wants to use all of the binders and pencils to create identical sets of office supplies for her classmates.During a car race, two competitors will stop to fill up the fuel tank.One does it every 4 laps, and the other every 5 laps. And immediately I say, well, I don't have to go any further.The least common multiple (LCM) of two or more positive integers is the smallest integer which is a multiple of all of them.And you see the point at which they have the same number is at 120. They both could have exactly 120 questions even though Luis's teacher is giving 30 at a time and even though William's teacher is giving 24 at a time. If you take a two away, you're not going to be divisible by 24 anymore. If you take a three or a five away, you're not going to be divisible by 30 anymore. And it's also dealing with dividing these things. So what's the largest number of prime numbers that are common to both factorizations? Then you don't have a three times anything else. So this is essentially telling us, look, we can divide both of these numbers into 3 and that will give us the largest number of identical sets. So we've answered the question is 3, but just to visualize it for this question, let's actually draw 21 binders. So if there are three people that are coming into this classroom, I could give them each seven binders and 10 pencils. And we essentially are just thinking about what's the number that we can divide both of these sets into evenly, the largest number that we can divide both of these sets into evenly.And so if you were to multiply all these out, this is 2 times 2 times 2 is 8 times 3 is 24 times 5 is 120. We want to divide these both into the greatest number of identical sets. 30 is, let's see, it's 3-- actually, I could write it this way-- it is 2 times 15. So let's say the 21 binders so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21. But that's the greatest number of identical sets Umama can make.

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