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I found this to be quite a problem! get involved

  1. Feb 14, 2012 #1
    Hey guys
    so yesterday the teacher gave us this problem and the one who knew how to solve it would win 5ive bucks ! somebody did eventually
    now, the guy being very odd didnt leak out any solution of his whatsoever ! i dont know the way of doing it so help me out id appreciate it
    ....
    HOW many natural numbers are out there between( 1...through...1000000), that are in the form of n^2, n^3 and n^4( of course n[itex]\in[/itex]N) ?
    12th grade discrete math

    Edit thanks to the warning : well now im thinking the max of all three terms is clear, then we have to just count them like:: n2 [itex]\rightarrow[/itex] ,1000 is the maximum so there are 1000 numbers available for that
    now all those numbers couldn't be applied for n3 and n4 because they would come out exceeding 1000000 !
    i need a proper way to count those numbers
     
    Last edited: Feb 14, 2012
  2. jcsd
  3. Feb 14, 2012 #2

    micromass

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    So the number must be of the form [itex]n^2[/itex] AND of the form [itex]n^3[/itex] AND of the form [itex]n^4[/itex]??

    Can you show that such a number must actually be of the form [itex]n^{12}[/itex]??
     
  4. Feb 14, 2012 #3
    the variable is "n", only those forms.. we can put any natural number in, for instance if we put 100 for n^2 we get 10000 but we can NOT exceed 1000 for n^2 because it would be more than 1 million
     
  5. Feb 14, 2012 #4
    anybody got an idea?
     
  6. Feb 14, 2012 #5

    Mark44

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    Do you understand what micromass is saying? What you wrote suggests that you don't. You're not putting a number in for n2 - you're putting a number (such as 1000) in for n, from which you see that n2 = 1,000,000.
     
  7. Feb 15, 2012 #6
    oh sorry my bad! i didn't read it properly !!!
    no im afraid we can't ! they are individual numbers not related in any ways, each has its own set of answers, but the final answer would be a sum of all 3hree sets !
     
  8. Feb 15, 2012 #7

    Mark44

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    Now you're not reading your own problem properly. Each number that you identify in the range 1 through 1,000,000 has to be a perfect square, and a perfect cube, and a perfect fourth power. There is no addition going on ("the final answer would be a sum of all 3hree sets").

    The only way that a given number satisfies all three conditions (perfect square, perfect cube, perfect fourth power) is for the number to be a perfect twelfth power. That's the essence of micromath's tip in post #2.
     
  9. Feb 15, 2012 #8
    aha now i see.
    so twelfth power would satisfies all three? for 4 nd 2 we take 4 andwe just multiply 4 and 3 to get 12 which is good for all !
    i got it :D thanks
     
  10. Feb 21, 2012 #9
    @mark44
    we actually have to count them...
    we are not allowed to sum'em up
    they are each different terms of this problem
    after we got the answers we do this: set of answer={n2[itex]\cap[/itex]n3 [itex]\cap[/itex] n4}
     
  11. Feb 21, 2012 #10

    Deveno

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    i think what you mean is this:

    how many numbers n are there, where:

    1 ≤ n ≤ 1,000,000

    n = x2 = y3 = z4.

    note that if n = w12, we can take x = w6, y = w4, and z = w3.

    on the other hand, if n = z4, then we can always take x = z2 (we get the fact that n is a square for free, since any 4-th power is also a square).

    that is: {n: n = z4, for some z in N} is a subset of {n: n = x2, for some x in Z}.

    and if A is contained in B, A∩B = A.

    so what we really need to show is THIS:

    if for two natural numbers y,z: y3= z4, z is a perfect cube. now z4 = (z2)2, so:

    √(y3) = z2

    and √(y3) = (√y)3 = y√y.

    since y√y = z2, which is a natural number, we must have that:

    √y = z2/y is a natural number, so y is a perfect square, say y = t2.

    so y√y = t3 = z2, so z2 is a perfect cube.

    again, repeating the same argument with:

    t√t = z, we see that t must be a perfect square, so t = s2, and thus z = s3, that is, z is a perfect cube.

    so all we need to do, is find all "perfect 12-th powers between 1 and 1,000,000" (find the "s's" or the "w's").

    312 = 531,441
    412 = 16,777,216, too big.

    thus a complete list is: {1, 4096, 531441}
     
  12. Feb 22, 2012 #11
    great explanation thanks but:
    im afraid N[itex]\neq[/itex]n2,n3,n4
    why you guys are insisting on this..
    n2= 1,...,1000 are eligible so we already have 1000 numbers to count for n squared
    for n3 couple more to count
    for n4 related to the n2 so we split them up
    so we'd have to get 1 thousand and some numbers in our answer set
     
  13. Feb 22, 2012 #12

    micromass

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    Aaah, I see what you mean. But you phrased the problem wrong.

    You want the numbers which are EITHER a square, EITHER a cube or EITHER a fourth power or a combination.

    So first, can you count all the squares (you already done this: there are 1000 numbers). Can you also count all the cubes and all the fourth powers??
     
  14. Feb 22, 2012 #13
    Bingo ..
    yeah sorry my bad, i was wrong at the first place
    exactly, we're sure about 1k of them,
    yes we can and we have to. The tricky part is how to find the forth and the third powers, perfect fourths are sort of squared powers, now we need 3hird powers as well !
     
  15. Feb 22, 2012 #14

    micromass

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    Maybe you can do something with [itex]\sqrt[3]{1000000}[/itex] and [itex]\sqrt[4]{1000000}[/itex]??
     
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