153 Lies Dormant in every positive multiple of 3
THEOREM:
Given a positive multiple of 3, sum the cubes of the digits.
Repeat this process. Eventually, you will get 153.
PROOF
Overview: Brute force for numbers up to 1000, and 4 particular 4-digit numbers.
All multiples of 3 bigger than 1000 (except for the 4 particular 4-digit numbers already
brute-forced), are bigger than the sum of their digits cubed. So, using induction, assume true
for all numbers less than x. Need to prove for x, but after one iteration
of the cubing process, you have a number less than x, so we're done.
Let x be a positive multiple of 3.
- CASE 1: x < 1000, or x = 1089, 1098, 1299, or 1899.
We can use brute force in, say, Excel, to check all the cases.
See gross attachment.
- CASE 2: x > 1000, but not equal to 1089, 1098, 1299, or 1899.
If x is a 4-digit number, then x is bigger than the sum of the cubes of
it's digits. (See Maple Code)
If x is an n-digit number, where n > 4, then the largest value that the first
iteration of summing the cubes can be is n 93, or 729 n. Furthermore,
x > 10n - 1 > 729n.
Thus, we have that for all x in this case, the first iteration is less than x itself.
Therefore, by induction, the theorem holds.