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 Interesting Function (2) (Posted on 2004-09-03)
Given: f is a function with domain and range of the positive integers, and f satisfies these two conditions:

(1) f(n+1) > f(n); that is, f is strictly increasing, and

(2) f(f(n)) = 3n

Find all possible values of f(955)

 See The Solution Submitted by SilverKnight Rating: 3.5000 (4 votes)

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 Solution | Comment 2 of 6 |

Let the monotonicity condition be Cond. 1 and let the condition f(f(n))=3n be Cond. 2.  Applying f to both sides of Cond. 2, we have

f(f(f(n)))=f(3n),

or applying Cond. 2 to that,

3f(n)=f(3n),

which we will call Cond. 3.

Let f(1)=a.  Then, applying Cond. 2 and 3, we have

f(a)=3
f(3)=3a

Now, since 3<=3a, we have by Cond. 1 that a<=3.  We cannot have a=3, since then we have 3=f(3)=9.  Also, we cannot have a=1, since then we have f(1)=3=f(3).  Thus, we have a=2, giving us the initial conditions

f(1)=2
f(2)=3
f(3)=6

Repeatedly applying Cond. 3 gives us:

f(3)=6
f(9)=18
f(27)=54
f(81)=162
f(243)=486
f(729)=1458

as well as

f(2)=3
f(6)=9
f(18)=27
f(54)=81
f(162)=243
f(486)=729
f(1458)=2187

Applying Cond. 1 to the equations f(729)=1458 and f(1458)=2187 gives us f(n)=n+729 for 729<=n<=1458, implying that f(955)=1684.

 Posted by David Shin on 2004-09-03 13:15:47

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