$97+98+ ...........+114+115 = 2014$. Here sum of $19$ consecutive numbers is $2014$. Find the largest number of consecutive positive integers whose sum is exactly 2014 and justify why you think this must be the largest number.

As far I did,
$\dfrac{n}{2}( 2a + n-1 ) = 2014$
$\implies 2an + n^2-n=4028$
Now we have to find the largest value of $n$.

Note This is a problem from BdMO 2014 National.

  • 3
    $\begingroup$ Hint: your expression factors as $n(n+2a-1)=4028$. It follows that $n$ is a divisor of $4028$. As $n+2a-1 ≥n$ we see that $n≤\sqrt {4028}$. $\endgroup$ – lulu Feb 1 '16 at 14:36

Let the first number in the sequence of positive integers be $m+1$ and the last number be $n$.


$$\dfrac{n(n+1)}{2}-\dfrac{m(m+1)}{2}=2014$$ $$n^2+n-m^2-m=4028$$ $$(n-m)(n+m)+(n-m)=4028$$ $$(n-m)(n+m+1)=4028$$

As $n$ and $m$ are positive integers, we have $n+m+1$ greater than $n-m$.



For the sequence to have maximum number of terms, we need $n-m$ to be a factor that is the highest possible while the other factor, $n+m+1$, is still greater than it. Moreover, for $n$ and $m$ to be integers, $n-m$ and $n+m+1$ must have opposite parity.

Thus $$n-m=53$$and $$n+m+1=76$$ On solving, $$n=64, m=11$$ Thus $$2014=12+13+\ldots+64$$ which is a sequence of $53$ terms.


Let's say $2014=a+(a+1)+\cdots+(a+k)$. If we denote by $m$ the average of this sequence, then we know that $2014=k\cdot m$. Now $m$ is either an integer or of the form $n+\frac{1}{2}$ for some integer $n$.

If $m$ is an integer, then $k$ is odd and we need $k\mid 2014$. Since $2014=2\cdot 19\cdot 53$ the only options for $k$ are $1$, $19$, $53$ and $19\cdot 53=1007$.

$k=1$ gives: $$ 2014=2014 $$ Which has length $1$.

$k=19$ gives the example sequence.

$k=53$ results in $m=38$ so: $$ 2014=12+\cdots+64 $$ Which has length $53$.

$k=1007$ results in: $$ 2014=-501+-500+\cdots+505=502+\cdots+505 $$ Which has length $4$.

Now if $m=n+\frac{1}{2}$ then $k$ is even, so $k=2q$ for some integer $q$. Now we have: $$ 2014=k\cdot m=2q\left(n+\frac{1}{2}\right)=q(2n+1) $$ So we are in the previous case after identifying $q\to m$ and $(2n+1)\to k$. "Expanding around" $n+\frac{1}{2}$ gives the same sequences as above when we vary $2n+1\in\{1,19,53,1007\}$.


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