#3**+10 **

Very nice, Melody......in a slightly different manner.....the sum of the first n integers =

[n (n + 1)] / 2 <= 250

n^2 + n <= 500

n^2 + n - 500 <= 0

On the graph here : https://www.desmos.com/calculator/ajecyupfie , the largest positve root = about 21.87....so, we can take the floor of 21.87 as the answer = 21 rows

CPhill
Jan 5, 2016

#2**+10 **

Hi Solveit and Guest ,

250 toothpicks how many rows (pyramid) could you make

1 on the top

2 on the next row

3 on the next row etc

\(1+2+3+4+...... +n \le 250\\~\\ \)

1+2+3+....+n

is the sum of an AP

\(a=1\;\;\;d=1 \;\;\;\\ Sn=\frac{n}{2}(a+L)\\ Sn=\frac{n}{2}(1+n )\\ so\\ \frac{n}{2}(1+n )\le250\\ n(1+n )\le500\\ n^2+n-500 \le0\\ \mbox{Solve for 0}\\ n=\frac{-1\pm\sqrt{1+2000}}{2}\\ n=\frac{-1\pm\sqrt{2001}}{2}\\ \mbox{n is not negative}\\ n=\frac{-1+\sqrt{2001}}{2}\\ n=21.8\\ \mbox{so there can be 21 rows} \)

231 toothpickswill be used andthere will be 19 left over :)

Melody
Jan 4, 2016

#3**+10 **

Best Answer

Very nice, Melody......in a slightly different manner.....the sum of the first n integers =

[n (n + 1)] / 2 <= 250

n^2 + n <= 500

n^2 + n - 500 <= 0

On the graph here : https://www.desmos.com/calculator/ajecyupfie , the largest positve root = about 21.87....so, we can take the floor of 21.87 as the answer = 21 rows

CPhill
Jan 5, 2016

#4**+5 **

Suppose the toothpicks are arranged as follows (i.e. 3 per row; after all they are long and thin!):

Then you get 12 rows with 16 left over.

Alan
Jan 5, 2016

#5**0 **

That is great Alan. I bet that is exactly what was intended.

I did think my pyramid was very weighed down (flattened). Maybe with time. LOL

Then again maybe my toothpicks were standing upright. That would have made a very tall skinny 'tower' I guess. :)

Melody
Jan 5, 2016

edited by
Guest
Jan 5, 2016