Determine the maximum possible value of a positive base 10 perfect cube M, such that we will obtain another perfect cube by deleting the last three digits of M.

Note: M cannot contain any leading zeroes and each of the last three digits of M is nonzero.

y=b^3-(10a)^3
y=(b-10a)(100a^2+10ab+b^2)
Since y is less than 1000, the only solutions are {a,b,y}={1,11,331}{1,12,728}
So M=1728=12^3

Posted by broll on 2011-10-15 11:02:11