# Puzzle Using Cubes

While playing one day, Megan decided to build cubes out of a box of individual sugar cubes. Megan emptied the box onto the floor and using the sugar cubes as building blocks, Megan made three solid cubes with no sugar cubes left over. Each of the cubes contained more than one sugar cube, and the three cubes may or may not have been the same size.

At this point, the family dog ran into the room and sent the cubes flying in all directions. The dog picked up one of the sugar cubes and left, crunching noisily.

Megan decided to start over, but this time she created two cubes instead of three. As before, each cube contained more than one sugar cube, the cubes may or may not have been the same size, and there were no sugar cubes left over. What is the smallest number of sugar cubes that could have been in the box when Megan began?

NOTE: I had no idea what to tag this post with, so I'm sorry If I got it seriously wrong.

• Hm, where I come from , "sugar cubes" are not actually cubes – Hagen von Eitzen Feb 12 at 0:33

An easy upper bound is $$1737$$ as $$1729=10^3+9^3=12^3+1$$ is well-known and we can add a $$2^3$$ cube on both sides. This is small enough to do a brute force search. It turns out that $$\tag1251=6^3+3^3+2^3=5^3+5^3+1$$ is the smallest solution. Indeed, if $$n=a^3+b^3+c^3=d^3+e^3+1^3$$ is a smaller solution, then $$(d,e)$$ must be one of $$(6,3)$$, $$(6,2)$$, $$(5,4)$$, $$(5,3)$$, $$(5,2)$$, $$(4,4)$$, $$(4,3)$$, $$(4,2)$$, $$(3,3)$$, $$(3,2)$$, $$(2,2)$$, and we can test if $$244$$, $$225$$, $$190$$, $$153$$, $$134$$, $$128$$, $$92$$, $$73$$, $$55$$, $$36$$,$$17$$ are sums of three cubes.