4 ms·
First solution (IMO 2002) is completely wrong. It shows that 1,2,3 cubes are not sufficient, and provide an obstacle that doesn't rule out 4 cubes, but does not
by tempfile 2y ago
First solution (IMO 2002) is completely wrong. It shows that 1,2,3 cubes are not sufficient, and provide an obstacle that doesn't rule out 4 cubes, but does not prove that there actually are 4 cubes that sum to the given number. This is much harder (and I don't know the true answer)
- kevinventullo 2y agoI think it’s about the same level of difficulty as showing you need at least four, but yeah I agree that what is written is insufficient. One solution is that you can write 2002 = 10^3 + 10^3 + 1^3 + 1^3 Then, multiply through by 2002^2001, which is itself a cube since 2001 is divisible by 3.
- tempfile 2y agoActually, I concede. I just didn't spot that.