
153
Explain this:
but first, notice this: if from 153 you take every digits (1,5,3) compute the cube of those digits and add them, you get back 153
(1^3 + 5^3 + 3^3) = 1 +125 + 27 = 153

Then pick any 3 digits number that can be divided by 3, any one
and do the same thing, then with the answer found, just repeat the same thing again, and again. You will always end up with 153
Example
take 342
3^3 + 4^3 + 2^3 = 99
9^3 + 9^3 =1458
1^3 + 4^3 +5^3 + 8^3 = 702
7^3 + 2^3 = 351
3^3 + 5^3 + 1^3 = 153
and other examples
666 gives you the sequence 648, 792, 1080, 513, 153
261 gives the sequence 225, 141, 66, 432, 99, 1458, 702, 351, 153
369 gives the sequence 972, 1080, 513, 153
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Re: 153
Let f(x) be the sum of the cubes of the digits of x. It can be shown that $\displaystyle x > f(x)$ for all $\displaystyle x \ge 1999$.
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 threedigit 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.