Let f(x) be the sum of the cubes of the digits of x. It can be shown that for all .

Therefore, if x is at least 1999, then iterating f several times, we will arrive at some number less than 1999, in which f must cycle around some number (this is true due to a Pigeonhole argument, if we iterate f 2000 times, at least two of the outputs must be the same).

It suffices to show that 153 is the only three-digit multiple of 3 that is equal to the sum of the cubes of its digits, which I'll leave that up to you. Apparently 370, 371, and 407 all share the same property, but none of them are multiples of 3.