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hana88

if x is an integer divisible by 3, then x^2 is divisible by 3. is this true or false? is the converse true or false?

  • one year ago
  • one year ago

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  1. hana88
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    can any one help?

    • one year ago
  2. just.chris
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    x = N x 3, represents the divisible by 3 integer. x^2 then equals N^2 x 3^2 which having the factors of 3 is therefore divisible by 3. Is this useful?

    • one year ago
  3. hana88
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    yes thank you

    • one year ago
  4. hana88
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    what about the converse?

    • one year ago
  5. lopus
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    X=2*3 6/3=2 \[x^{2}=36\] 36/3=12

    • one year ago
  6. hana88
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    that only gives me evidence

    • one year ago
  7. just.chris
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    I haven't done this stuff in awhile. Do you mean if and integer, x, is divisible by 2 then x^3 is also divisible by 2?

    • one year ago
  8. hana88
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    no if x^2 is divisive by 3, then x is divisible by 3

    • one year ago
  9. just.chris
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    OK. I am winging it here In order for x^2 to be divisible by a prime, which cannot be further factored then it must also be a factor of x. Is this useful?

    • one year ago
  10. lopus
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    x=n*3 n=1,2,3,4,5,6,7,8............... x^2=3,36,81,144....... divide

    • one year ago
  11. hana88
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    @lopus x is 3,6,9,12,18...

    • one year ago
  12. just.chris
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    x = 3,6,9,... is correct valus for x but does that indicate that if x^2 is divisible by 3 then so is x? I am asking because I don't know. A question about the question.

    • one year ago
  13. lopus
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    yes, you just divide and see for yourself demostration x=N*3 (N*3)^2 =n^2 * 3^3

    • one year ago
  14. lopus
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    excuse me 3^2

    • one year ago
  15. just.chris
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    So you don't really need the series you just need if x^2 is divisible by 3 then x is because 3 cannot be further factored. And the equations you provided apply in bothe directions. unlike say if x^2 is divisible by 4 then x is not necessarily divisible by for because 4 can be further factored, i.e., 2^2 = 4 is divisible by 4 but x, being 2, is not. Yes?

    • one year ago
  16. Mertsj
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    If x^2 is divisible by 3 it can be written in the form 3a where a is an integer. So x^2=3a and x = sqrt3(sqrt a) so the converse is not true.

    • one year ago
  17. helder_edwin
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    3 is a prime number. u should pay attention to that!!!

    • one year ago
  18. hana88
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    um @helder you make no sense

    • one year ago
  19. Mertsj
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    Why?

    • one year ago
  20. hana88
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    @Mertsj thanks thats the best answer because it was easy to understand

    • one year ago
  21. helder_edwin
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    if \(3\mid x^2\) then \(3\mid x\) because 3 is prime.

    • one year ago
  22. Mertsj
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    yw

    • one year ago
  23. Mertsj
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    That is not true helder. 3|15 but 3 does not divide sqrt15

    • one year ago
  24. Mertsj
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    So if x^2 = 15, 3 is a factor of x^2 but not a factor of x

    • one year ago
  25. helder_edwin
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    but 15 is not a perfect square!

    • one year ago
  26. Mertsj
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    The converse is: If x^2 is divisible by 3 then x is an integer divisible by 3. So we see that x^2 does not have to be a perfect square. It only has to be divisible by 3.

    • one year ago
  27. helder_edwin
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    so why it is x^2 if it is not a perfect square?

    • one year ago
  28. Mertsj
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    x^2 could be 15 or 18 or 21 or 24 or 27 or 12 .

    • one year ago
  29. helder_edwin
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    nerver mind.

    • one year ago
  30. just.chris
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    If x^2 does not have to be a perfect square then the converse is not true. However, if x^2 is perfect and divisible by 3 then is the converse true? And would that be the case for any prime? Since primes cannot be factored into smaller integers then I think in order for the square to be factorable by the prime then you must have the prime^2 in order to have the prime as a factor of x^2.

    • one year ago
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