Number property, page 65, #15

[cristhian]

Hello, I'm having difficulty understanding the answer to this question: if x/y has a remainder of 0 and z/y has a remainder of 3, what is the remainder of xz/y. The book says the remainder is 0 since x/y is 0, but z/y has a remainder of 3, so I don't understand why this doesn't come into play.

Thank you!
Melissa

[yx33]

Look at it this way:

since x/y has 0 remainder, x/y = an integer, for simplicity let's call that integer (^_^), or smiley face.

since xz/y = (x/y)*(z)
therefore, xz/y = (^_^)*z

Since this has become a multiplication problem between 2 integers, it will naturally have no remainders.

[cristhian]

Thank you for this explanation! I don't understand why you would multiple the Z after. If you were actually doing this problem, wouldn't you multiple Z times X and then divide by y?For example: 2*3/4 (or whatever numbers we use). It seems like it should be x/y(z) for you to multiple it after, right?

Thank you!

[yx33]

Hey, Melissa! It doesn't matter in what order you calculate the x, y and z, the final answer is the same. The order only matters when you have "+" and/or "-" mixed in with multiplication and division.
To put it in another way:
(xz)/y = x*(z/y)= (x/y)*z

[tommywallach]

Great conversation here. I would explain it a little more simply:

If x/y has a remainder of zero, than x is a multiple of y.

No matter what you multiply x by, it will STILL be a multiple of y (by the very definition of MULTIPLE!), so xz is a multiple of y, meaning there's no remainder.

Hope that helps!

-t

[cristhian]

Thank you both - this helped a lot!

[tommywallach]

Glad to help!

-t