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B A question regarding multiples of 3

  1. Dec 20, 2016 #1
    Why doesn't any odd multiple of 3 can give "24-1(x2-1)=y" as a result (y) a natural number? Obviously no even number will make "y" a natural number, but all of the odd numbers do, but odd multiples of 3 (3, 9, 15, 21, 27...).

    x=1⇒y=0
    x=3⇒y=1/3
    x=5⇒y=1
    x=7⇒y=2
    x=9⇒y=10/3
    x=11⇒y=5
    x=13⇒y=7
    x=15⇒y=28/3
    x=17⇒y=12
    x=19⇒y=15
    x=21⇒y=55/3
    .
    .
    .
     
    Last edited: Dec 20, 2016
  2. jcsd
  3. Dec 20, 2016 #2

    jedishrfu

    Staff: Mentor

    Can you explain this better perhaps by using an example?
     
  4. Dec 20, 2016 #3
    I'm sorry, I typed it wrongly, it's not "1/2", it's "1/24".
     
  5. Dec 20, 2016 #4

    jedishrfu

    Staff: Mentor

    Okay, but can you explain this better perhaps by using an example?

    Is this a homework assignment?
     
  6. Dec 20, 2016 #5
    There, now I think it's WAY better understandable. I'm sorry. Lol
     
  7. Dec 20, 2016 #6

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    You mean, why does the expression:
    $$\frac{1}{24}(x^2-1)$$
    not return an integer when ##x## is an odd multiple of 3? Think about what "odd multiple of 3" means mathematically and substitute that for ##x## in the expression above to see what you get.
     
  8. Dec 20, 2016 #7

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Can x2-1 be divisible by 3 if x is divisible by 3?
     
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