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Seeking direction

  1. Jun 26, 2004 #1
    hello all. i am in dire need of direction here and i really appreciate all this help.

    how many 0's are in 1100!

    prove tuv=(tu,tv,uv)[t,u,v]

    only prime that makes 4p+1 a perfect square is p=2.

    can anyone please help me? i am in so much need of direction... :confused: :cry:
     
  2. jcsd
  3. Jun 26, 2004 #2

    matt grime

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    for the one about highest common factors, this time I would in the first instance just prove it using prime decomposition, though that is not the most admirable way.

    suppose 4p+1=x**2, rearrange and factorize and use the defining properties of primes in Z: a number, q, is prime iff when ever q|ab then q|a or q|b.

    Remember, this is just mathematics, things just follow from the definitions, sometimes by a clever trick but not usually in questions like this.

    Yes there is still soem work for you to do here, but you need to indicate how you're trying to solve them, so start with these hints and tell us how far you get.
     
  4. Jun 26, 2004 #3

    Gokul43201

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    1. What contributes a '0' to the decimal representation of a product of numbers ? Think about 5s and 2s. Also think about what is the highest power,k, of a prime number, p < n, such that p^k divides n! Try some examples and you'll find a pattern...
     
  5. Jun 26, 2004 #4
    :smile: woooo, it is by these people on this great site that is making me understand this theory stuff!!! God love you all!!! :smile:
     
  6. Jun 27, 2004 #5
    perhaps there are 189 zeros?
     
  7. Jun 27, 2004 #6
    matt, by setting it up like this, do i solve for the p b/c that is what i am looking for right? do i'd have 4p+1=x**2. = 4p=x^2-1 and then divide by 4, thus, p=x^2-1/4. pick an x and show that it is prime? reply and let me nkow if this is right.

    i'm sorry gokul, i just cannot find a pattern here yet... i can see how i.e. p=11 but that is all i can come up with.. i'm sorry if i'm not seeing it yet plese maybe a little more insight??
     
  8. Jun 29, 2004 #7

    matt grime

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    Please don't divide like that. Use divisibilty. Since 4p=(x+1)(x-1) and prime factorizations are unique in Z then one of the following cases must hold: x+1 is one of 1,2,4,p,2p or 4p, and x-1 the other factor. Which of those systems gives of equations is solvable in Z with p prime?
     
  9. Jun 29, 2004 #8

    Gokul43201

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    How many multiples of 5 among [1,1100] ? Clearly, this is = floor(1100/5) = 220.

    But then, every multiple of 25 gives you and extra factor of 5 not counted above. There are floor(1100/25)=44 such 5s.

    Similarly there are floor(1100/125) = 8 additional factors of 5 from multiples of 125.

    And finally, there is 625, which gives you another extra 5.

    So the higest power of 5 in 1100! is 220+44+8+1=273 {which is just = floor(1100/5) + floor(1100/5^2) + floor(1100/5^3) + floor (1100/5^4) }

    Similarly, you can find the highest power of 2 in 1100! The lower of these two numbers gives you the highest power of 10 in 1100!, which is the number of zeros.
     
  10. Jun 29, 2004 #9

    Gokul43201

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    Loop Quantum Gravity,

    One of us is wrong. I think I'm right.
     
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