This isn't a homework problem but I found it on the internet and can't figure out how to do it. It's one of those "divisible by whatever"-type problems which I never learned how to solve...

I don't know how to work with this divisibility stuff when it's generalized to numbers like x+y+z. I don't even know where to begin. All I can do is "9w = x+y+z for some number w"...and now I don't know where to go. If I think up example values for x,y, and z I find that (a) keeps working, but I don't know why. Any help appreciated!

100x + 10y + z + 9= 99x+ x+ 9y+ y+ z+ 9= 9(11x+ y+ 1)+ x+y+z
Does that answer the question for the first?

"reduce modulo 9" mean look for the part this isn't divisble by 9- the remainder when you divide by 9.

100= 9(11)+ 1 so 100 reduces to 1 modulo 9 and so 100x reduces to x modulo 9
10= 9+ 1 so 10 also reduces to 1 modulo 9 so 10y reduces to 7 modulo 9
Obviously 1 reduce to 1 modulo 9 so z reduces to z modulo 9
9 itself reduces to 0 modulo 9
In other words 100x+10y+z+ 9 consists of stuff that is obviously divisible by 9 plus x+ y+ z, the "modulo 9" part.

in more layman terms, all you are doing is more or less finding remainders when dividing by 9. writing numbers N as N = 9d +r, where r is the remainder, d is some integer. For x+y+z, you have x+y+z = 9d +0. One thing you do not know from the question however is whether x,y,z are all individually divisible by 9. But if so, i guess that would make the question trivial given those 5 choices available.

I think he is trying to solve a problem from an SAT work book which includes trigonometric, algebraic and geometric arithmetic and not algebra (abstract algebra) so he might not be familar with the Modulo Arithmetic system? I had no idea what it was until I worked through a few chapters of a few different modern abstract algebra texts.

If he isn't, I don't know if he is going to understand it without a more clear definition, then again, I guess I can't make a judgment claim about a person I have never met. I have a slow learning curve so it might just be me.