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Sal solves the following challenge: Given the positive integers a, b, and c, where a is a multiple of c and (a+b)/c is an integer. Is b necessarily a multiple of c? Created by Sal Khan.
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- can you separate a+b+d/c into a/c+b/c+d/c?(6 votes)
- Yes, if they are within a parenthesis, then absolutely. However, if these conditions aren't met, then you try to solve as is... the second format that you're describing is associated with solving for one within variable solutions with an algebraic answer that most likely would be long but simple to solve. Also, the MOST important thing of all is to make sure that c cannot = 0; if it does then you tore a rift in reality...(15 votes)
- At the very end how does he know that just because b is an integer means that it's a multiple of c?? Im so lost(5 votes)
- Can we multiply an integer?(5 votes)
- The way I did it was this:
I wrote the expression a/c = n, where n represents an integer, then I changed it into a = nc and substituted it back into (a+b)/c, which gave me (nc+b)/c.
Then I simplified this into (nc)/c + b/c = n + b/c, and since n is an integer b/c must also be an integer.
Is this an alright way to do it?(7 votes)
- That's interesting - you sort of did the opposite of Sal's method. (Sal's way is how I've always thought about these fractions.) But this seems like a perfectly valid way to look at it.(0 votes)
- Hello, How are you? Thanks a lot for your videos. They're awesome and helpful.
I just want to ask you about the b/c thing. If:
a+b/c will equal 3
a is a multiple of/ and divisible by c
b is not a multiple of c, but it can be divided by c, and it's an integer.
The result of a+b/c is an integer, too.
Am I wrong?(4 votes)
- You are correct. Kinda. You had the theory correct, just not the math. When you use order of operations, you can divide b by c which is 2/2 which is simply 1. So it's going to like like:
4+2/2=4+1/1. Then, it's going to look like 4+1 which is five. If you had done a-b/c, then you would get 3. Hope this helped!(3 votes)
- If you have a multiple of a certain number and add that to another multiple of that number you get a multiple of that number.
In other words ax+bx=(a+b)x
When you have multiple numbers over the same denominator you do the operations in the numerator first. So in this case you would do a+b first.
Because a is divisible by c, for (a+b)/c which is the same as a/c + b/c to be an integer than b would have to be divisible by c right?(2 votes)
- Yes, if a/c is an integer and a/c + b/c is also an integer then b/c is an integer.(2 votes)
- It seems like we're conflating being a multiple of c with being an integer. If a is 4 and c is 2, b could be 0. here, b is an integer, (a + b)/c is an integer (it's 2). However, is 0 a multiple of c? That doesn't seem right, but perhaps it is. What's the formal definition for "being a multiple of"?(2 votes)
- In the example you've provided your forgetting the first bit of information that we were provided with which is that b>0, therefore b cannot be 0.(2 votes)
- what does he mean when he say that a is both divisible and a multiple of c?(2 votes)
- Those are two different words to describe the same relationship between numbers. The connotations are slightly different, but when one is true, the other will be as well.(1 vote)
- What is a multiple again? I know I'm weird.(2 votes)
- A multiple is a number that can be produced from taking a number and multiplying it by another number. To think of it in another sense, if a number, let's call it n, is a true multiple of another, let's call it x, then n can be divided by x without any remainder; and you can take x and multiply it by another number to get n. For example: 10, 25, 30, 35, and 40 are all multiples of 5 because they can be divided by 5 without any remainder, and you can take 5 and multiply by some other number, and get 10, 25, 30, 35, and 40. Did that help?(1 vote)
- Hello there! I was doing the exercise after watching this video and I encountered this problem:
There are B boys and 8 girls in a classroom, and there are more boys than girls. Each student in the classroom has at least 2 pencils. There are P pencils in the classroom in total.
Is the following statement true?
P > 30
The answer was yes.
The total number of students in the classroom is (b+8)
If they each have at least 2 pencils, then there will be at least 2(b+8) pencils or:
If p>32, then it must also be true that p>30, so the statement is true.
I don't understand how the part why p>2(8+8). Where does the first "8" come from? I can understand for the number of girls, but where did the other one come from? Please help! Thanks! :-)(1 vote)
- There are 8 (this is the first 8) girls in the classroom.
There are more boys than girls.
That means FOR SURE there are 8 (this is the second 8) boys in the classroom - we are told there are more, but not how many more, but that doesn't matter. If there are more boys than girls, and there are 8 (1st eight) girls, then there are at least 8 (2nd eight) boys.(2 votes)
Let's say that we have three integers, a, b, and c, and we know that all of these integers are greater than 0. So they're integers, and they are greater than 0. And we also know that the expression a plus b over c, that this is also an integer. The entire expression, if you were to evaluate it, is also an integer. And then finally, we know that a is divisible-- or another way of saying it, that a is a multiple of c, so a is divisible by c, which is another way of saying is a is a multiple of c. So this is what we know. a, b, and c are integers, all greater than 0. We know that the expression a plus b over c is also an integer, and that a is a multiple of c, or a another way is that c divides perfectly into a. So our question for you or the question for all of us to work out right now is, is b a multiple? Does b have to be a multiple of c? Let me write it that way. Does b-- given all of these constraints-- does b have to be a multiple of c? So let's see how we can-- and I encourage you to pause the video right now to come up with your own answer about whether b has to be a multiple of c. So now that you've unpaused things, let's work it out. So let's go to our original expression right over here. We have a plus b over c, and really one way to tackle this, is to really just play around with this expression, and see if we can come up with any conclusions here. So, one, we could try to rewrite a plus b over c. We could rewrite that as a/c plus b/c, and this expression is the exact same thing as our first expression. So we know that this entire thing is going to be an integer. That whole thing is going to be an integer. Now, what do we know about these parts? Well a/c, this is a divided by c. We know that a is divisible by c. We know that a is a multiple of c. So divided by c, this is going to be an integer, so let me write that. So, this information right over here tells us that this thing right over here-- that a divided by c-- is going to be an integer. This is going to be an integer. Now, if I have an integer, and I add something to it, and the whole thing is an integer, well the thing that I'm adding to it must be an integer. The only way that I get an integer plus something to be an integer is if the thing I'm adding it to is also an integer. So this-- there's no way that I could add an integer to a non-integer and then get an integer, so this has to be an integer. And if b/c is an integer, that means that, yes, b must be a multiple of c. So the answer here is yes.