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anonymous

  • one year ago

someone could help me here , please .

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    i don't know what method i have to use there. :l

  3. UsukiDoll
    • one year ago
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    I only have a slight idea for the first one given n is an odd integer \[\geq 5 \] so your n must be an odd number greater than or equal to 5

  4. UsukiDoll
    • one year ago
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    x, y - positive integers.. so we just have to pick all positive numbers

  5. UsukiDoll
    • one year ago
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    @baad1994 try letting x = 1 , y = 3, and n = 7 for x+2y=n

  6. perl
    • one year ago
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    *

  7. UsukiDoll
    • one year ago
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    @perl I have no idea if I had the right idea for the first one.... T__________________T I think I did, but it's only some combinations

  8. dumbcow
    • one year ago
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    since there are an infinite num of odd integers, there will be an infinite num of (x,y) pairs i guess they want an expression in terms of n first notice that x = n - 2y , so 2y <= n-1 since x must be positive also y is pos integer 1,2,3 ... n-1 then num of possible y-values and num of pairs for any given n is (n-1)/2 total num of pairs is then just the infinite sum \[\sum_{i=2}^{\infty} \frac{n-1}{2}, n = 2i +1\] which simplifies to \[\sum_{i=2}^{\infty} i\]

  9. perl
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @dumbcow since there are an infinite num of odd integers, there will be an infinite num of (x,y) pairs i guess they want an expression in terms of n first notice that x = n - 2y , so 2y <= n-1 since x must be positive `also y is pos integer 1,2,3 ... n-1` \(\color{blue}{\text{End of Quote}}\) Question. Should that say y is a pos. integer 1,2,3,.. (n-1)/2

  10. dumbcow
    • one year ago
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    ahh yes, thank you for the correction

  11. perl
    • one year ago
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    Nice solution :)

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spraguer (Moderator)
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