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## Precalculus

### Course: Precalculus > Unit 4

Lesson 5: Modeling with rational functions- Analyzing structure word problem: pet store (1 of 2)
- Analyzing structure word problem: pet store (2 of 2)
- Combining mixtures example
- Rational equations word problem: combined rates
- Rational equations word problem: combined rates (example 2)
- Mixtures and combined rates word problems
- Rational equations word problem: eliminating solutions
- Reasoning about unknown variables
- Reasoning about unknown variables: divisibility
- Structure in rational expression

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# Reasoning about unknown variables: divisibility

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.

## Want to join the conversation?

- 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)
- He says that b/c is an integer. For b/c to be an integer c has to divide into b evenly.(13 votes)

- Can we multiply an integer?(5 votes)
- Yes. an integer is just any whole number that is not zero.(4 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=4

b=2

c=2

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)

- Hello,

I fail to see how those activities relate to rational functions.(2 votes)- These might not be the rational expressions you are used to seeing. But they still involve variables in the denominator, so they are still rational.(1 vote)

- 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)

## Video transcript

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.