Divisibility by 9: Solving x+y+z Problems

In summary, 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...
  • #1
sh86
19
0
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...

Given x, y and z such that x+y+z is divisible by 9, which of the following must also be divisible by 9?

a) 100x + 10y + z + 9
b) 10y + z + 9
c) 9x + 90y + z + 9
d) x + y + 9z + 9
e) 100x + 10y + 7z

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!
 
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  • #2
Reduce the expression modulo 9.
 
  • #3
I don't know what that means.. I don't think anything big is supposed to be used on this test since I found it on a SAT practice website.
 
  • #4
Modular arithmetic isn't hard. It's fun! But alternatively, write a) as:

99*x+9*y+9+x+y+z

Now everything is divisible by 9, right?
 
  • #5
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.
 
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  • #6
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.
 
  • #7
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.

Post number [tex]10^2[/tex]!

Sorry, I am bored at work.
 
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1. What is the rule for determining if a number is divisible by 9?

The rule for determining divisibility by 9 is to add up all the digits in the number. If the sum is divisible by 9, then the original number is also divisible by 9.

2. Can numbers with more than 3 digits also be tested for divisibility by 9 using this rule?

Yes, the rule for divisibility by 9 can be applied to numbers with any number of digits. Simply add up all the digits until you are left with a single digit number, then check if that number is divisible by 9.

3. How does this rule apply to solving x+y+z problems?

When solving x+y+z problems, the rule for divisibility by 9 can be used to check if the sum of the three numbers is divisible by 9. If it is, then the original problem can be simplified by dividing each number by 9.

4. Can the rule for divisibility by 9 be applied to fractions or decimals?

Yes, the rule can be applied to fractions and decimals by treating the digits after the decimal point as separate numbers. For example, if the number is 3.45, you would add 3+4+5=12, and 12 is divisible by 9 so 3.45 is also divisible by 9.

5. Are there any other ways to determine if a number is divisible by 9?

Another way to determine divisibility by 9 is to use the divisibility rule for 3, which states that if the sum of the digits in a number is divisible by 3, then the number is also divisible by 3. Since 9 is a multiple of 3, this rule can also be applied to determine divisibility by 9.

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