ganeshie8
  • ganeshie8
Are there any positive integers that cannot be expressed as sum of at least two consecutive integers ?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.