+0

# toothpicks

+3
763
5

250 toothpicks how many rows (pyramid) could you make

Jan 4, 2016

#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   Jan 5, 2016

#1
0

i think 250/8 = 31.25  (31)

Jan 4, 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 :)

Jan 4, 2016
#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
#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.

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. :)

Jan 5, 2016
edited by Guest  Jan 5, 2016