Volume question

Question

A box holds small cube shaped blocks that are the same size. Kim tires tk build a large cube out of the small blocks but finds that she needs 6 more blocks. Takashi builds a different sized cube out of the same amount of blocks and finds that she has 85 blocks left. How many blocks are in the box?

Answer 1

Let Kim be building a cube of side $x$ and Takashi be building a cube of side $y$ where $y<x$.

So $x^3 -6 = y^3 + 85$

$x^3 - y^3 = 91$

$(x-y)(x^2 + xy +y^2) = (7)(13) = (1)(91)$

where the RHS has been factorised in the only two unique ways among natural numbers.

Now try $x-y = 7 \implies x = y + 7$ and then $x-y = 1 \implies x = y+1$, substitute those into the equation and solve the resulting respective quadratics arising from the second factor on the LHS. You'll find the only admissible solution to be $x = 6, y = 5$, so the number of blocks is $210$.

Answer 2

Say there are $k$ blocks in the box.

Say Kim tries to build a cube of side $n$, so: $n^3 = k+6$

Say Takashi tries to build a cube of side $m$, so: $m^3= k-85$

Where $k, m, n$ are all integers. Oh, and we can see: $n^3-m^3 = 91$