On a large piece of paper, Dana creates a "rectangular spiral" by drawing line segments of lengths, in cm, of $1,1,2,2,3,3,4,4, \ldots$ as shown. Dana's pen runs out of ink after the total of all the lengths he has drawn is $3000 \mathrm{~cm}$. What is the length of the longest line segment that Dana draws?
(A) 38
(D) 55
(B) 39
(E) 30
(C) 54

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-106.jpg?height=336&width=331&top_left_y=1727&top_left_x=1409)
On a large piece of paper, Dana creates a "rectangular spiral" by drawing line segments of lengths, in cm, of $1,1,2,2,3,3,4,4, \ldots$ as shown. Dana's pen runs out of ink after the total of all the lengths he has drawn is $3000 \mathrm{~cm}$. What is the length of the longest line segment that Dana draws?
(A) 38
(D) 55
(B) 39
(C) 54

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-464.jpg?height=328&width=347&top_left_y=1294&top_left_x=1363)

## Solution

The formula for the sum of the natural numbers from 1 to $n$ is $\frac{(n)(n+1)}{2}$.

That is, $1+2+3+\ldots+n=\frac{(n)(n+1)}{2}$.

(For example, $1+2+3+\ldots+10=\frac{(10)(11)}{2}=55$.)

We can find the sum of a double series, like the one given, by doubling each side of the given formula.

We know $1+2+3+\ldots+n=\frac{(n)(n+1)}{2}$.

If we double each side we get $2(1+2+\ldots+n)=n(n+1)$.

So, $(1+1)+(2+2)+(3+3)+\ldots+(n+n)=n(n+1)$.

In this question we want the value of $n$ so that the following is true:

$(1+1)+(2+2)+(3+3)+\ldots+(n+n) \leq 3000$.

Or, if we use the formula $(n)(n+1) \leq 3000$.

We would now like to find the largest value of $n$ for which this is true.

The best way to start is by taking $\sqrt{3000} \doteq 54.7$ as a beginning point.

If we try $n=54$, we find $(54)(55)=29703000$. So $n=55$ is not acceptable.)

This means that $(1+1)+(2+2)+(3+3)+\ldots+(54+54)=2970$ so that the longest length that Dana completed was $54 \mathrm{~cm}$. (If we had included the length 55 then we would have had a sum of 3025 which is too large.)

ANSWER: (C)