The function F(x) is defined for all positive real values of x
and is such that:
F(3x) = 3F(x), and:
F(x) = 1 - abs(x-2) for 1 ≤ x ≤ 3.
Determine the minimum value of x for which F(x) = F(2001).
*** abs(x) refers to the
Let T be the largest power of 3 less than X. That is, T = 3 ^ floor(log_3(X)).
Then F(X) = F(X/T) * T
= [1 - abs(X/T - 2)] * T
= T - abs(X - 2T)
If (X-2T) >= 0, then F(X) = 3T-X.
If (X-2T) < 0, then F(X) = X-T.
In other words, F(X) is the difference between X and its nearest power of 3.
For X = 2001, T = 729, X-2T > 0 so F(X) = 3*729 - 2001 = 186.
Quick observation finds that the smallest positive X which is exactly 186 away from it's nearest power of 3 is 429.
For X = 429, T = 243, x-2T < 0 so F(X) = 429 - 243 = 186.
Posted by tomarken
on 2014-11-20 08:55:55