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ganeshie8

  • one year ago

Are there any positive integers that cannot be expressed as sum of at least two consecutive integers ?

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  1. ganeshie8
    • one year ago
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    2 = not possible 3 = 1+2 4 = not possible 5 = 2+3 6 = 1+2+3 7 = 3+4 ...

  2. omarbirjas
    • one year ago
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    8

  3. ganeshie8
    • one year ago
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    yes 8 also cannot be expressed as sum of consecutive integers

  4. ParthKohli
    • one year ago
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    Let the first number be \(p\) and let there be \(q\) consecutive integers.\[k = pq + \dfrac{q(q-1)}{2}\]

  5. omarbirjas
    • one year ago
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    to tired to do any more math...

  6. ParthKohli
    • one year ago
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    I don't know how recasting the problem into this form helps, but let's see.

  7. ganeshie8
    • one year ago
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    that looks good because it gives us something to play with

  8. ganeshie8
    • one year ago
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    k = p + (p+1) + (p+2) + ... + (p+q-1) = pq + q(q-1)/2 so k must be of this form is it

  9. ParthKohli
    • one year ago
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    exactly.

  10. ParthKohli
    • one year ago
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    Umm, I haven't even answered the question...

  11. ganeshie8
    • one year ago
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    we're 90% done actually

  12. ganeshie8
    • one year ago
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    \[k = pq + \dfrac{q(q-1)}{2}\] multiplying 2 through out gives \[2k = q(2p+q-1)\] clearly the factors on right hand side cannot have same parity (cannot be both odd or both even)

  13. ganeshie8
    • one year ago
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    thats because the sum of factors : \(q+2p+q-1 = 2(p+q)-1\) is odd

  14. ganeshie8
    • one year ago
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    so \(2k=q(2p+q-1)\) is impossible when \(k\) is a power of \(2\)

  15. anonymous
    • one year ago
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    can someone help me with trigonometric raios? sin(3x+13) = cos(4x)

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