ganeshie8
  • ganeshie8
Are there any positive integers that cannot be expressed as sum of at least two consecutive integers ?
Mathematics
jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
2 = not possible 3 = 1+2 4 = not possible 5 = 2+3 6 = 1+2+3 7 = 3+4 ...
omarbirjas
  • omarbirjas
8
ganeshie8
  • ganeshie8
yes 8 also cannot be expressed as sum of consecutive integers

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ParthKohli
  • ParthKohli
Let the first number be \(p\) and let there be \(q\) consecutive integers.\[k = pq + \dfrac{q(q-1)}{2}\]
omarbirjas
  • omarbirjas
to tired to do any more math...
ParthKohli
  • ParthKohli
I don't know how recasting the problem into this form helps, but let's see.
ganeshie8
  • ganeshie8
that looks good because it gives us something to play with
ganeshie8
  • ganeshie8
k = p + (p+1) + (p+2) + ... + (p+q-1) = pq + q(q-1)/2 so k must be of this form is it
ParthKohli
  • ParthKohli
exactly.
ParthKohli
  • ParthKohli
Umm, I haven't even answered the question...
ganeshie8
  • ganeshie8
we're 90% done actually
ganeshie8
  • ganeshie8
\[k = pq + \dfrac{q(q-1)}{2}\] 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)
ganeshie8
  • ganeshie8
thats because the sum of factors : \(q+2p+q-1 = 2(p+q)-1\) is odd
ganeshie8
  • ganeshie8
so \(2k=q(2p+q-1)\) is impossible when \(k\) is a power of \(2\)
anonymous
  • anonymous
can someone help me with trigonometric raios? sin(3x+13) = cos(4x)

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