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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?
• 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...
• 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
• He says that b/c is an integer. For b/c to be an integer c has to divide into b evenly.
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?
• 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.
• Can we multiply an integer?
• Yes. an integer is just any whole number that is not zero.
• Hello, How are you? Thanks a lot for your videos. They're awesome and helpful.
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?
• 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!
• 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"?
• 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.
• 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?
• Yes, if a/c is an integer and a/c + b/c is also an integer then b/c is an integer.
• Hello,
I fail to see how those activities relate to rational functions.
• 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?