 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Minimum Value Muse 2 (Posted on 2014-11-20) 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 absolute value of x.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 2

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 Please log in:

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