[Prof. Templeton]

Long before Professor Templeton started his career at Redrum University he had a job as a stockboy at the Manhattan Fruit Exchange. One morning a large pallet of oranges arrived that Prof. Templeton was tasked with stacking into a display. When the oranges arrived they were laid out in a perfect square on the pallet. When the Prof. was done he had stacked them all into a four sided pyramid with none left over. Wiping the sweat from his brow, the Prof. marvelled at the fact that he had used all the oranges when making his pyramid and he wondered how that was possible. He knew that there had to have been over 3,000 oranges but surely no more than 6,000. How many oranges, exactly, had arrived to be stacked that morning?

[HoustonHokie]

The professor-to-be stocked 4900 oranges that day.

[Guest]

But I'm going to write the answer anyway:

4900

By trial and error*. Is there a clever way to come to the answer?

*Or comparing and contrasting square and pyramidical numbers.

[HoustonHokie]

4900 is 70², and is also the summation of the series O=X², where O equals the number of oranges on a particular level, and X is the level, starting at the top of the pyramid. It does take a little trial and error to find the right value of X (which is 24), but that's the most elegant way I found to solve it. Just a minute or so in a spreadsheet to get the answer.

I note that the professor's pyramid was stable, as he smartly placed oranges in upper levels into the spaces between oranges in the lower levels. Question: does an orange at the bottom level get crushed by all the weight of the oranges above? There are, after all, 24 levels of oranges, which means the bottom 576 oranges are supporting the weight of 4324 oranges above them. Sounds like it could be orange juice .

By the way, did you know that my Hokies are Orange Bowl bound?

[Guest]

4900 is 70², and is also the summation of the series O=X², where O equals the number of oranges on a particular level, and X is the level, starting at the top of the pyramid. It does take a little trial and error to find the right value of X (which is 24), but that's the most elegant way I found to solve it. Just a minute or so in a spreadsheet to get the answer.

I note that the professor's pyramid was stable, as he smartly placed oranges in upper levels into the spaces between oranges in the lower levels. Question: does an orange at the bottom level get crushed by all the weight of the oranges above? There are, after all, 24 levels of oranges, which means the bottom 576 oranges are supporting the weight of 4324 oranges above them. Sounds like it could be orange juice .

By the way, did you know that my Hokies are Orange Bowl bound?

I knew that !!

[Prof. Templeton]

4900 is 70², and is also the summation of the series O=X², where O equals the number of oranges on a particular level, and X is the level, starting at the top of the pyramid. It does take a little trial and error to find the right value of X (which is 24), but that's the most elegant way I found to solve it. Just a minute or so in a spreadsheet to get the answer.

I note that the professor's pyramid was stable, as he smartly placed oranges in upper levels into the spaces between oranges in the lower levels. Question: does an orange at the bottom level get crushed by all the weight of the oranges above? There are, after all, 24 levels of oranges, which means the bottom 576 oranges are supporting the weight of 4324 oranges above them. Sounds like it could be orange juice .

By the way, did you know that my Hokies are Orange Bowl bound?

Correct! 4900 oranges. Manhattan Fruit Exchange uses little cardboard stackers to keep the friut from squishing.

I had come across this puzzle in a different form (cannonballs I think) so I knew an answer existed, but had forgotten the actual numbers. I used trial and error myself, although the numbers are listed on Mathworld and you can find some formulas on wikipedia. Congrats to VT, may they stomp the Bearcats.

Side-puzzle: If the oranges had arrived in a perfect cube (any quantity except 1 of course), could a square pyramid have been created with no left-overs?

[Guest]

Side-puzzle: If the oranges had arrived in a perfect cube (any quantity except 1 of course), could a square pyramid have been created with no left-overs?

I'd been wondering about this one myself.

then with 13 million or fewer oranges, the answer would be no (I haven't dragged my calculation past a 340-orange-side pyramid base)

However, I don't think the oranges would have been stacked in a cube that way

to balance an orange on top of an orange. Wouldn't each row fill in the gaps from the previous row, like your pyramid did? I haven't checked this one arrangement out yet.

Edited December 29, 2008 by Cherry Lane

[Guest]

I'd been wondering about this one myself.

However, I don't think the oranges would have been stacked in a cube that way

to balance an orange on top of an orange. Wouldn't each row fill in the gaps from the previous row, like your pyramid did? I haven't checked this one arrangement out yet.

then with 13 million or fewer oranges, the answer would be no (I haven't dragged my calculation past a 340-orange-side pyramid base)

where there is essentially a cube of side n-1 inside a cube of side n, it works when n=4 (that's 91 oranges, so the pyramid would have a base of 6 oranges per side) It would't work within your original constraint of 3000-6000 oranges

[Guest]

I don't think you need to say that there are less than 6,000 oranges.

The next closest answer would require 10,200,725,887,841,500 oranges which would be humanly impossible to stack into a pyramid.

[HoustonHokie]

I tried a couple other combinations where the base didn't increase by 1 orange, but by 2, 3, 4, or more (1, 16, 49, ...; 1, 25, 81, ...; 1, 36, 121, ...; and so on) and didn't find a solution for less than 1,000,000,000,000,000 oranges (floating point operations gave out at this point, which is why I didn't get edgukated's answer) until I tried the following:

1² = 1

24² = 576

47² = 2209

70² = 4900

93² = 8649

116² = 13456

total oranges = 29791 = 31³

For this solution, the dimension of the base increased by 23 oranges with each level.

A pretty squat pyramid, but a square pyramid nonetheless.

I imagine there are other possibilities with pyramids like this one, and maybe with some that don't come to a 1 orange peak (maybe a 4 orange or 9 orange peak would be better). As long as the base increased by more than the dimension of the top level each time, I think they'd still have some validity as square pyramids and may be worth consideration.

[bonanova]

they arrived in a 70x70 square and ended up in a pyramid with 24x24 base.

Total is 4900 oranges.

Nice puzzle

For the cube, I checked up to 5000 x 5000 [pyramid] base without solution.

[Prof. Templeton]

Can't be done. I came across a neat little proof while researching this puzzle. I had originally thought the puzzle was taking a cube and making a pyramid, but then realized my mistake and it started with a square. I like CL's interpretation of the nesting oranges. Nice.

[Guest]

Well it can't be done if one of the oranges is not exactly cube-shaped which is quite rare.

[Prof. Templeton]

Well it can't be done if one of the oranges is not exactly cube-shaped which is quite rare.

Well, I kind of meant cubed in a case of some sort.

[Guest]

Answetr is 4900 it's simple.