2021 Mathcounts practice competition target #4 question #2

Question

What is the greatest number of positive consecutive numbers that sum to 400?

My approach:

We have terms from : $(x-n)..... x.....(x+n)$
This would mean the sum of these terms is
$(2n+1) * 2x * 1/2 = x(2n+1) = 400$
$2n+1$ MUST be an odd number, so $x$ is even
$400 = 20^2 = 2^4 * 5^2$
The odd multiple is either $5$, with $x$ being 80, or we have $25*16$ &25*16& would have more terms. Since x is 16, we know we have the range (x-12) = 4 to (x+12) = 28
Therefore, we have 25 terms as the maximum.

The correct answer is 27, and I don't know how to get that. I considered the possibility of an even amount of terms , but 27 isn't even so that wouldn't matter. Where did I go wrong / Is the real answer wrong?

PROOF THIS IS NOT AN ONGOING TEST:

I have the real answer.

Answer 1

Let the sum be $\sum_{k=1}^n (x+k)$, where $x\geqslant 0$. This sum has $n$ positive consecutive terms and equals

$$nx + (1+2+\dots+n) = nx + n(n+1)/2 = \frac n2(2x+n+1)$$

In other words, we need to find the maximal integer $n$ such that for some integral $x\geqslant 0$ we have

$$\frac n2(2x+n+1) = 400 \iff n\big(2x+(n+1)\big) = 800$$

From this, we have that $n$ divides $800$, so $27$ cannot be the answer.
Of course, $n\big(2x+(n+1)\big) > n^2$ so $n$ must be less than $\sqrt{800} \simeq 28.28$, which implies $n\leqslant 28$.

Now, $26$, $27$ and $28$ do not divide $800$, so our first candidate is $25$. We get

$$25(2x+26) = 800 \implies 2x + 26 = 32 \implies x = 3,$$

which confirms $n=25$ as being the correct answer.

Answer 2

Let $400=(x+1)+(x+2)+\cdots+(y-1)+y$, where $y\ge x>0$. Now, we have $400=\frac{x(x+1)}2-\frac{y(y+1)}2$, or equivalently, $800=(x-y)(x+y+1)$. Thus, we wish to find a factorization $800=n\cdot m$ where $m$ and $n$ have different parities, $m>n\ge0$, and which maximizes $n$ (the length of the sum). After some experimentation we see that $n=25$ and $m=32$ is the best possible.