Yet another interesting math puzzle...?

Question

In a snowball fight, where snowballs are identical spheres, your opponents have stacked their snowballs in a square pyramid. You are about to count the snowballs along the bottom edge of the opponent's stack when one appears with another snowball. After giving him a telling off, the opposition's leader takes apart the square pyramid and builds a new, triangular pyramid using all the original snowballs and the extra one. Find two possible values for the number of snowballs your opponents now have.

Perhaps someone can come up with an equation to solve this? If not, the raw answers would still be helpful. Have fun...

Answer 1

The number must be the sum of squares + 1 and the sum of triagular numbers.
So SUM (i=1 to n) i^2 +1= n(n+1)(2n+1)/6 +1 is the number in the first.
BUT the triangular one is SUM (i=1 to m) i(i+1)/2 =
1/2 * SUM (1 to m) i^2+i = m(m+1)(2m+1)/12+ m(m+1)/4

Now find m,n such that the two expressions are equal. Start with values for m, find the height of a triagular pyramid of height m, see if there is a corresponding one of height n.

Another answer here assumed the pyramids were the same height (both n). This is not correct.

Good luck!

Answer 2

Let there be n stacks in the square pyramid

So, in bottom stck there are n^2

in the upper stack there are (n-1)^2

So, total no. is 1+2^2 + 3^3 + ............+n^2 = n(n+1)(2n-1)/6

Now, in triangular one,

nth stack contains n(n+1)/2

So, total is

1+ 2(2+1)/2 + 3(3+1)/2 + ............ + n(n+1)/2

= 1/2 [ n(n+1)(2n-1)/6 + n(n+1)/2]

Answer 3

56 or 286.

It's pretty easy to model this with an Excel spreadsheet.

The sequence of square pyramids:
1,5,14,30,55,91,140,204,285,385,506,650,819,...

The sequence of triangular pyramids:
1,4,10,20,35,56,84,120,165,220,286,364,455,...

They could have gone from 55 to 56, or from 285 to 286.

Answer 4

you are a math guru, you must be knowing this