Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - Help with number problem

  1. #1
    Junior Member
    Joined
    Mar 2007
    Posts
    33

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Unt0t View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2007
    Posts
    33
    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
    Last edited by Unt0t; March 14th 2007 at 08:32 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Unt0t View Post
    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.

    there's something i dont like about this proof, i'll think about it and get back to you
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Unt0t View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Mar 2007
    Posts
    33
    Umm I just don't understand the 3|x, 4|x etc. bit... like what does " | " actually mean?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Unt0t View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Jhevon View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2007
    Posts
    33
    Quote Originally Posted by Jhevon View Post
    do you know of logic functions, like => or <=> or that stuff?
    Are those ment to be greater/less equal signs?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Mar 2007
    Posts
    33
    Quote Originally Posted by Jhevon View Post
    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.

    there's something i dont like about this proof, i'll think about it and get back to you
    Now I understand what you've done here.... but are you sure that the second question is asking that?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Unt0t View Post
    Are those ment to be greater/less equal signs?

    no. what is the highest level of math that you have done?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Mar 2007
    Posts
    33
    ???
    Last edited by Unt0t; March 14th 2007 at 08:56 PM.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Unt0t View Post
    7th form maths
    what does that mean?? have you taken calculus or any proof class?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Mar 2007
    Posts
    33
    What ???
    Last edited by Unt0t; March 14th 2007 at 08:57 PM.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Unt0t View Post

    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.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Birthday Problem; Prime Number Problem
    Posted in the Algebra Forum
    Replies: 9
    Last Post: January 8th 2012, 12:35 PM
  2. Replies: 2
    Last Post: December 18th 2008, 06:28 PM
  3. number problem
    Posted in the Algebra Forum
    Replies: 6
    Last Post: June 1st 2008, 07:40 PM
  4. Replies: 6
    Last Post: March 30th 2008, 11:39 AM
  5. number problem
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: December 23rd 2007, 08:58 PM

Search Tags


/mathhelpforum @mathhelpforum