SOLUTION: Clara organizes cans in triangular piles, where each row has one less can than the row below. For example, the pile of 15 cans has 5 cans in the bottom row and 4 cans in the row a

Algebra ->  Trigonometry-basics -> SOLUTION: Clara organizes cans in triangular piles, where each row has one less can than the row below. For example, the pile of 15 cans has 5 cans in the bottom row and 4 cans in the row a      Log On


   

Question 1150514: Clara organizes cans in triangular piles, where each row has one less can than the row below. For example, the pile of 15 cans has 5 cans in the bottom row and 4 cans in the row above it.
(a) There are 3240 cans in a pile. How many cans are in the bottom row?
(b) There are S cans and they are organized in a triangular pile with 'n' cans in the bottom row. Show that n^2 +n - 2S= 0
(c) Clara has 2100 cans. Explain why she cant organize them in a triangular pile.
Found 2 solutions by greenestamps, htmentor:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The number of cans in a pile with n cans on the bottom row is

S = 1 + 2 + 3 + ... + (n-1) + n

The sum of the integers from 1 to n is

S+=+%28n%28n%2B1%29%29%2F2

(a) S = 3240

3240+=+%28n%28n%2B1%29%29%2F2
6480+=+n%28n%2B1%29

Solve by inspection: 80(81) = 6480, so n = 80.

(b) n%5E2%2Bn-2S+=+n%5E2%2Bn-2%28%28n%28n%2B1%29%29%2F2%29+=+n%5E2%2Bn-%28n%5E2%2Bn%29+=+0

(c) S = 2100

2100+=+%28n%28n%2B1%29%29%2F2
4200+=+n%28n%2B1%29

The solution to that equation is not an integer, so there can't be that many cans in the pile.

Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
a) This problem involves an arithmetic sequence with common difference of -1, and the last term equal to 1.
The nth term of an arithmetic sequence is a_n = a_1 + (n-1)d
So we have 1 = a_1 + (1-n)
Thus a_1 = n
The sum of an arithmetic sequence with n terms is
S_n = (n/2)(a_1 + a_n)
6480 = a_1^2 + a_1
The solutions are a_1 = 80, -81
Take the positive solution, a_1 = 80
b) The above equation can be written 2S = n((n + 1) -> n^2 + n - 2S = 0
c) n^2 + n - 2*2100 = 0
n^2 + n - 4200 = 0
There are no integer solutions, thus this number of cans cannot be made into a triangular pile