# Help with number problem

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• Mar 6th 2007, 06:35 PM
Unt0t
Help with number problem
Okay I got this number problem and I don't really know what the questions are trying to say.

1) Give sufficient examples to convince someone that every multiple of 3 that is divisible by 4 is also a multiple of 6.

2) Show algebraically that this is true in general for any such numbers.

Any help would be greatly appriciated.
• Mar 6th 2007, 07:01 PM
Jhevon
Quote:

Originally Posted by Unt0t
Okay I got this number problem and I don't really know what the questions are trying to say.

1) Give sufficient examples to convince someone that every multiple of 3 that is divisible by 4 is also a multiple of 6.

2) Show algebraically that this is true in general for any such numbers.

Any help would be greatly appriciated.

what math is this? what level of math have you done? i want to know what kind of solution to give
• Mar 6th 2007, 07:06 PM
Unt0t
yeah and I have no idea what i'm ment to give as an answear... Like I know how to do everything in that course, except this question is worded quite weirdly
• Mar 6th 2007, 07:15 PM
Jhevon
Quote:

Originally Posted by Unt0t
Okay I got this number problem and I don't really know what the questions are trying to say.

1) Give sufficient examples to convince someone that every multiple of 3 that is divisible by 4 is also a multiple of 6.

2) Show algebraically that this is true in general for any such numbers.

Any help would be greatly appriciated.

1) will five examples be enough?

24, 36, 72, 144, 216

2) Let x be a multiple of 3. then 3|x. (3|x means "3 divides x," that is 3 goes into x and leaves no remainder.)

since 3|x, x = 3m for some integer m.

similarly, if x is also a multiple of 4, then 4|x. so x = 4n for some integer n
=> 3m = 4n
=> 3m/4 = n, where n is some integer divisible by 3 and 4.

Now suppose 6|n, then n = 6k for some integer k. that is 3m/4 = 6k.
so 3m = 24k = 6(4k).
since 4k is an integer, and 3m = 6(4k), 6|3m. but 3m is x, and so 6|x, and therefore, x is a multiple of 6 as well.

• Mar 6th 2007, 07:24 PM
Jhevon
Quote:

Originally Posted by Unt0t
This is a quesion that is part of my assignment I got for my Maths 102, it's like a basic maths paper for first year university. yeah and I have no idea what i'm ment to give as an answear... Like I know how to do everything in that course, except this question is worded quite weirdly

does this stuff look familiar to you?
• Mar 6th 2007, 07:28 PM
Unt0t
Umm I just don't understand the 3|x, 4|x etc. bit... like what does " | " actually mean?
• Mar 6th 2007, 07:45 PM
Jhevon
Quote:

Originally Posted by Unt0t
Umm I just don't understand the 3|x, 4|x etc. bit... like what does " | " actually mean?

do you know of logic functions, like => or <=> or that stuff?
• Mar 6th 2007, 07:46 PM
Jhevon
Quote:

Originally Posted by Jhevon
do you know of logic functions, like => or <=> or that stuff?

"|" means "divide". if i say 3|9 it means 3 divides 9, which is true, dividing 9 by 3 leaves no remainder when done. neverthe less i dont want to use anything you are not familiar with. tell me what you have done in class.
• Mar 6th 2007, 07:52 PM
Unt0t
Quote:

Originally Posted by Jhevon
do you know of logic functions, like => or <=> or that stuff?

Are those ment to be greater/less equal signs?
• Mar 6th 2007, 08:11 PM
Unt0t
Quote:

Originally Posted by Jhevon
1) will five examples be enough?

24, 36, 72, 144, 216

2) Let x be a multiple of 3. then 3|x. (3|x means "3 divides x," that is 3 goes into x and leaves no remainder.)

since 3|x, x = 3m for some integer m.

similarly, if x is also a multiple of 4, then 4|x. so x = 4n for some integer n
=> 3m = 4n
=> 3m/4 = n, where n is some integer divisible by 3 and 4.

Now suppose 6|n, then n = 6k for some integer k. that is 3m/4 = 6k.
so 3m = 24k = 6(4k).
since 4k is an integer, and 3m = 6(4k), 6|3m. but 3m is x, and so 6|x, and therefore, x is a multiple of 6 as well.

Now I understand what you've done here.... but are you sure that the second question is asking that?
• Mar 6th 2007, 08:11 PM
Jhevon
Quote:

Originally Posted by Unt0t
Are those ment to be greater/less equal signs?

no. what is the highest level of math that you have done?
• Mar 6th 2007, 08:14 PM
Unt0t
???
• Mar 6th 2007, 08:27 PM
Jhevon
Quote:

Originally Posted by Unt0t
7th form maths

what does that mean?? have you taken calculus or any proof class?
• Mar 6th 2007, 08:30 PM
Unt0t
What ???
• Mar 6th 2007, 08:50 PM
Jhevon
Quote:

Originally Posted by Unt0t

2) Show algebraically that this is true in general for any such numbers.

Any help would be greatly appriciated.

here is a better proof, i think.

Background knowledge:

you should know that an integer is even if you can express it as 2n for some integer n, it is odd if you can express it as 2n + 1 for some integer n. if a number is divisible by 6 (a multiple of 6), you can write it as 6n for n an integer, if its divisible by 4, you can write it as 4n for n an integer and so on.

I'll try to tone done the logic and formal math for this proof.

We want to show that if a number x is divisible by 4 and divisible by 3 then it is divisible by 6.

Proof: assume that x is not divisible by 6. then we can write x as 6n + r, where n is any integer and r is the remainder when dividing by 6. therefore, r = 1,2,3,4, or 5.

and so we have 5 cases.

case 1: x = 6n + 1 (that is when we divide x by 6 we have 1 as a remainder).

note that x = 6n + 1 = 3(2n) + 1, since 2n is an integer, it means x is not divisible by 3 (we have a remainder 1).

case 2: x = 6n + 2

notice x = 6n + 2 = 3(2n) + 2
that means we have 2 as a remainder when we divide x by 3, so x is not divisible by 3

case 3: x = 6n + 3

note that x = 6n + 3 = 3(2n + 1), so in this case x is divisible by 3. now we will check if it is divisible by 4.

notice that x = 6n + 3 = 2(3n + 1) + 1. this means x is odd, and therefore is not divisible by 4.

case 4: x = 6n + 4

so x = 6n + 4 = 3(2n + 1) + 1, so x is not divisible by 3

case 5: x = 6n + 5

so x = 6n + 5 = 3(2n + 1) + 2, so x is not divisible by 3.

so we see in all cases that when a number is not divisible by 6 it is not divisible by either 3 or 4 as well. so we can conclude the opposite is true, by way of the contropositive.
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