CF 293C - Cube Problem
The three original cubes had side lengths $a$, $b$, and $c$. Their total number of unit cubes was $$a^3+b^3+c^3.$$ Vitaly wanted to build one larger cube whose side length was $a+b+c$, which would require $$(a+b+c)^3$$ unit cubes.
Rating: 2400
Tags: brute force, math, number theory
Solve time: 2m 37s
Verified: no
Solution
Problem Understanding
The three original cubes had side lengths $a$, $b$, and $c$. Their total number of unit cubes was
$$a^3+b^3+c^3.$$
Vitaly wanted to build one larger cube whose side length was $a+b+c$, which would require
$$(a+b+c)^3$$
unit cubes.
He was short by exactly $n$ cubes, so
$$(a+b+c)^3-(a^3+b^3+c^3)=n.$$
We are given only $n$, and we must count how many positive integer triples $(a,b,c)$ could have produced that value.
The bound $n \le 10^{14}$ immediately rules out any search over possible values of $a$, $b$, or $c$. Cube roots of $10^{14}$ are around $4.6 \times 10^4$, so even a triple loop over all possible side lengths would require more than $10^{13}$ iterations.
The key observation is that the expression has a strong algebraic structure. Once it is transformed into a multiplicative equation, the problem becomes one of enumerating divisors of a number up to about $3.3 \times 10^{13}$. Numbers of that size have relatively few divisors, making divisor-based enumeration practical.
Several edge cases are easy to mishandle.
For example, if
n = 1
then
$$(a+b+c)^3-(a^3+b^3+c^3)$$
is always divisible by $3$, so the answer must be $0$. A solution that starts enumerating divisors without checking divisibility by $3$ wastes work and may produce incorrect results.
Another subtle case is
n = 24
which corresponds to
$$3(a+b)(b+c)(c+a)=24.$$
The factor triple $(1,1,8)$ satisfies the product condition, but it does not correspond to positive integer values of $a,b,c$. Recovering $a,b,c$ requires both parity and positivity constraints. Ignoring those constraints overcounts.
A third common mistake is forgetting that $(a,b,c)$ is ordered. The triples $(1,2,3)$ and $(2,1,3)$ are different solutions. When we enumerate sorted factor triples, we must restore the correct multiplicity.
Approaches
Start from the identity
$$(a+b+c)^3-(a^3+b^3+c^3) = 3(a+b)(b+c)(c+a).$$
Let
$$x=a+b,\quad y=b+c,\quad z=c+a.$$
Then the problem becomes
$$3xyz=n.$$
If $n$ is not divisible by $3$, there are no solutions.
A brute-force idea would be to enumerate all factor triples $(x,y,z)$ whose product is $n/3$, reconstruct $a,b,c$, and count the valid ones. This is already much better than enumerating $a,b,c$, because the number of divisors of a $10^{13}$-sized integer is small.
The remaining challenge is counting efficiently.
Given
$$m=\frac n3,$$
we need all positive integer triples satisfying
$$xyz=m.$$
From
$$a=\frac{x+z-y}{2},\quad b=\frac{x+y-z}{2},\quad c=\frac{y+z-x}{2},$$
a triple corresponds to valid positive integers exactly when
$$x+y+z \equiv 0 \pmod 2$$
and the strict triangle inequalities hold:
$$x+y>z,\quad y+z>x,\quad z+x>y.$$
Since we enumerate factor triples in sorted order $x \le y \le z$, only
$$x+y>z$$
needs to be checked.
The brute-force factor enumeration is correct, but enumerating all positive triples independently would still be too large. The observation that $xyz=m$ lets us enumerate only divisors of $m$. Even for the worst possible $m$, the divisor count is only a few thousand, so checking divisor pairs is easily fast enough.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force over $a,b,c$ | $O((\sqrt[3]{n})^3)$ | $O(1)$ | Too slow |
| Divisor Enumeration | $O(d(m)^2)$ | $O(d(m))$ | Accepted |
Here $d(m)$ denotes the number of divisors of $m$.
Algorithm Walkthrough
- Read $n$.