Class 6 math (India)
GCF & LCM word problems
Here we have a couple of word problems--one searching for the least common multiple and the other for the greatest common factor. Just read them with us slowly and follow along. You'll get it. Created by Sal Khan.
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- Isn't the LCM of 8 and 6 going to be 24?
8*1 = 8 6*1 = 6 6*4 = 24
8*2 = 16 6*2 = 12
8*3 = 24 6*3 = 18(44 votes)
- Yes it is becuase
6x1=6 6x2=12 6x3=18 6x4=24
8x1=8 8x2=16 8x3=24
so it is 24(10 votes)
- Can HCF or LCM be negative?(27 votes)
- No, HCF and LCM need to be positive at all times.(9 votes)
- What is the different between gcf and lcm(11 votes)
- Factors and multiples are opposites of each other.
Multiples of 15 are: 15, 30, 45, 60, 75, etc.
Factors of 15 are: 1, 3, 5, and 15
Now, how do these apply to GCF and LCM.
Let's find the GCF of 15 and 9
Prime factors of 15 = 3 * 5
Prime factors of 9 = 3 * 3
The 2 numbers share one common factor. The GCF = 3.
Now, let's find the LCM for 15 and 9
Multiples of 15: 15, 30, 45, 60, 75, ...
Multiples of 9: 9, 18, 27, 36, 45, 54, 63, ...
The first common multiple (LCM) = 45
Hope this helps.(35 votes)
- "Umaima just bought 1 package of 21 binders. She also bought 1 package of 30 pencils. She wants to use all the binders and pencils to create identical sets of office supplies for her classmates."
Wouldn't the greatest number of sets be 21? 1 binder with 1 pencil in each set? Yes there would be pencils left over but who cares?(17 votes)
- I agree with your premise but what matters is what the problem asked for. It does say "She wants to use ALL the binders and pencils".(16 votes)
- At3:03, Sal says that the LCM can be calculated using prime factorization, and I am fluent in the method. I do not use the first method employed in the video's first problem.
So can I rely on prime factorization for all LCM problems in my test (which is an online multiple-choice test, where the method of calculation doesn't matter as I'll only calculate on rough sheets of paper)?(3 votes)
- Yes, you could use prime factorization for all LCM problems, and it would always work if you use the method correctly. The method involves using each prime factor the greatest number of times it occurs in any of the prime factorizations.
However, for some problems, this method is not always the most efficient method. Efficiency might matter if your test is timed. For example, if one of two numbers is a multiple of the other number, then the LCM of the two numbers is the larger number (for example, because 24 is a multiple of 8, LCM(8,24) is 24). Also, if two numbers have only 1 as a common factor, then the LCM of the two numbers is their product (for example, LCM(9,10) is 9*10=90, because the only common factor of 9 and 10 is 1).(9 votes)
- Why is it so HARD? and compicated(6 votes)
- Because you don't undestand it very well. Keep working on it and suddenly you'll reilize how simple it really is.(1 vote)
- What's the main difference between gcf and Lcm?(2 votes)
- Here's an example that might clarify things for you.
Let's find the common multiple of 6 and 4
Multiples of 6 are: 6, 12, 18, 24, 30, ...
Multiples of 4 are: 4, 8, 12, 16, 20, 24, ...
The LCM = 12. This is the 1st multiple that they have in common.
You may also see this referred to as LCD = Lowest Common Denominator.
Now, look at common factors.
Factors of 6: 1, 2, 3, 6
Factors of 4: 1, 2, 4
A lowest common factor would = 1 for all numbers.
The Greatest Common Factor (GCF) or Greatest Common Divisor (GCD) = 2. This is the largest number that you can divide evenly into both 4 and 6.
Hope this helps.(8 votes)
- When can you tell when it's going to be GCF or LCM?(5 votes)
- i don't get the video(3 votes)
- Factors are For example what can go into nine? Well 9x1 and that is one, 3 goes itno nine beacause 3x3 is 9. so our little chart is 1,3,9,, lets just stop there okay! Now to make it simple for the next thing which is you guessed it commen multipules, is pretty simple it is like a times table chart so you would go 9 x1= 9 2x9=18 3x9=27 And so on forth, lets take a look at what we have for cm well 9,18,27. Her eis a trick to remeber the nines
I so hope this helps!(4 votes)
- I am so confused! I got the problems in the video OK, but when I did the practice, I couldn't get it. Could someone explain please? Thx(4 votes)
- I did not understand the practice either.(2 votes)
William and Luis are in different physics classes at Santa Rita. Luis's teacher always gives exams with 30 questions on them, while William's teacher gives more frequent exams with only 24 questions. Luis's teacher also assigns three projects per year. 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. What is the minimum number of exam questions William's or Luis's class can expect to get in a given year? So let's think about what's happening. So if we think about Luis's teacher who gives 30 questions per test, so after the first test, he would have done 30 questions. So this is 0 right over here. Then after the second test, he would have done 60. Then after the third test, he would have done 90. And after the fourth test, he would have done 120. And after the fifth test, if there is a fifth test, he would do-- so this is if they have that many tests-- he would get to 150 total questions. And we could keep going on and on looking at all the multiples of 30. So this is probably a hint of what we're thinking about. We're looking at multiples of the numbers. We want the minimum multiples or the least multiple. So that's with Luis. Well what's going on with William? Will William's teacher, after the first test, they're going to get to 24 questions. Then they're going to get to 48 after the second test. Then they're going to get to 72 after the third test. Then they're going to get to 96. I'm just taking multiples of 24. They're going to get to 96 after the fourth test. And then after the fifth test, they're going to get to 120. And if there's a sixth test, then they would get to 144. And we could keep going on and on in there. But let's see what they're asking us. What is the minimum number of exam questions William's or Luis's class can expect to get in a year? Well the minimum number is the point at which they've gotten the same number of exam questions, despite the fact that the tests had a different number of items. And you see the point at which they have the same number is at 120. This happens 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. And so the answer is 120. And notice, they had a different number of exams. Luis had one, two, three, four exams while William would have to have one, two, three, four, five exams. But that gets them both to 120 total questions. Now thinking of it in terms of some of the math notation or the least common multiple notation we've seen before, this is really asking us what is the least common multiple of 30 and 24. And that least common multiple is equal to 120. Now there's other ways that you can find the least common multiple other than just looking at the multiples like this. You could look at it through prime factorization. 30 is 2 times 15, which is 3 times 5. So we could say that 30 is equal to 2 times 3 times 5. And 24-- that's a different color than that blue-- 24 is equal to 2 times 12. 12 is equal to 2 times 6. 6 is equal to 2 times 3. So 24 is equal to 2 times 2 times 2 times 3. 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. If it's going to be divisible by 30, it's going to have to have 2 times 3 times 5 in its prime factorization. That is essentially 30. So this makes it divisible by 30. And say, well in order to be divisible by 24, its prime factorization is going to need 3 twos and a 3. Well we already have 1 three. And we already have 1 two, so we just need 2 more twos. So 2 times 2. So this makes it-- let me scroll up a little bit-- this right over here makes it divisible by 24. And so this is essentially the prime factorization of the least common multiple of 30 and 24. You take any one of these numbers away, you are no longer going to be divisible by one of these two numbers. If you take a two away, you're not going to be divisible by 24 anymore. If you take a two or a three away. If you take a three or a five away, you're not going to be divisible by 30 anymore. 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. Now let's do one more of these. Umama just bought one package of 21 binders. Let me write that number down. 21 binders. She also bought a package of 30 pencils. She wants to use all of the binders and pencils to create identical sets of office supplies for her classmates. What is the greatest number of identical sets Umama can make using all the supplies? So the fact that we're talking about greatest is clue that it's probably going to be dealing with greatest common divisors. And it's also dealing with dividing these things. We want to divide these both into the greatest number of identical sets. So there's a couple of ways we could think about it. Let's think about what the greatest common divisor of both these numbers are. Or I could even say the greatest common factor. The greatest common divisor of 21 and 30. So what's the largest number that divides into both of them? So we could go with the prime factor. We could list all of their normal factors and see what is the greatest common one. Or we could look at the prime factorization. So let's just do the prime factorization method. So 21 is the same thing as 3 times 7. These are both prime numbers. 30 is, let's see, it's 3-- actually, I could write it this way-- it is 2 times 15. We already did it actually just now. And 15 is 3 times 5. So what's the largest number of prime numbers that are common to both factorizations? Well you only have a three right over here. Then you don't have a three times anything else. So this is just going to be equal to 3. 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 just to be clear of what we're doing. So we've answered the question is 3, but just to visualize it for this question, let's actually draw 21 binders. 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. And then 30 pencils, so I'll just do those in green. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Let me just copy and paste that. This is getting tedious. So copy and paste. So that's 20 and then paste that is 30. Now, we figured out that 3 is the largest number that divides into both of these evenly. So I can divide both of these into groups of 3. So for the binders, I could do it into three groups of 7. And then for the pencils, I could do it into three groups of 10. So if there are three people that are coming into this classroom, I could give them each seven binders and 10 pencils. But that's the greatest number of identical sets Umama can make. I would have three sets. Each set would have 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.